# Tag Info

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firstly V=IR is not ohm's law. it is the definition of resistance. ohms law states that V is proportional to I where R is a constant for ohmic conductors. in a voltmeter a very tiny current flows through a large resistance and a much larger current flows through the device which has a smaller resistance such that V=iR=Ir for ideal voltmeters the R tends to ...

2

It is possible to have a perfect voltmeter: You can use a potentiometer with a current meter but you also need something with a standard voltage (standard cell, whose emf is larger than what you are trying to measure). It is a null method, that works by varying the sliding contact on the potentiometer until a zero current is registered. Zero current means no ...

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An ideal voltmeter is made from a galvanometer connected to a resistor ,whose resistance is considered to be infinitely big, in series. But in practical life you can never achieve this because ....well... you can never reach infinity.

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If you argue that capacitors in parallel have the same p.d. across them and in series otherwise, then when the switch is thrown the capacitors are in series (the capacitors have different pd's) and the charged capacitor acts like a source of emf. After some finite time (assuming some finite residual resistance) afterwards the capacitors have the same pd so ...

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simple answer: They are in parallel when both blades of the knife switch are closed. If only one blade was closed, the 2 capacitors would be in series between the 2 open points.

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I would assume that entropy would be a large factor. In any conversion of energy from one form to another, there will naturally be some degree of waste heat (entropy) generated. Therefore, heating is like rolling something downhill- it's the "natural" tendency anyway. Cooling something generally involves moving heat up a gradient, which requires more energy ...

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Because potential exactly means that you have its potential at every point and to get from start to finish you need to make a number of steps. That is one leap can be broken into a series of steps, every step is closer to the target. As you move closer to the target, your potential becomes closer to the target potential. That is, you have some potential ...

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A lot of these problems are best done by first redrawing the circuit so that it is in a more accessible form. The correct answer is $\frac 8 3 \;\mu$F.

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Comment and Hint: The following figure may help, as long as you maintain all the connections this is the same circuit (will let you identify the capacitors). I get $8/3 \mu F$

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You can solve this strictly with adding capacitance in parallel and elastance in series. The 2uF is in parallel with the others. The two 1uF on the left are in parallel and the 1uF on the right is in series with the two on the left.

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With two lamps in series you will get double overall resistance, thus of Ohm's law overall current is I = U / (2R) that means overall current gets half of one-lamp case with I = U / R. As you said, this current is the same in the both lamps and voltage on each of the lamps gets half of the overall. In parallel you have two lamps attached to one the same ...

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A typical voltmeter contains an internal Ohmic resistor with known and very high resistance $R$ (called the "input resistance" or "input impedance"), and an extremely sensitive ammeter that measures the current through that resistor. When the voltmeter is connected in parallel across some circuit elements, then ideally the internal resistor has resistance ...

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The component that ensures the current is zero just after the switch is closed is the inductor. Inductors do not like changes in current, since a change in current means the magnetic field linking the inductor is changing and this generates a back emf that opposes the change. If you replace the inductor with a piece of wire you would have an RC circuit and ...

1

Setting up the differential equation $$L \frac {dI}{dt} + RI + \frac {Q}{C} = V$$ will not necessarily answer your question, "What and how can I conclude about the current in this circuit just after switch is closed." If you look at the methods of solving the differential equation somewhere on the way to the solution initial conditions are needed, one of ...

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Let say the power produced by a transmission company is 50 Giga_Watt(we can take any value really). What happens is, they convert this to very high voltage.When you use a 1 kilo_ohm resistor, the current draw should be 22 amps, but the power generated(50Kw) must be kept in mind! You cannot have a power source of 10 W and expect it to provide 10 Volts at 2 ...

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Answered my own question... My solution was two metal items in place of cups

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To question 2: When the electron reaches the end of a conductor, it would have to move into the air, which is an isolator. The entire conductor is at equal potential, which is much much lower than the potential at a point out in the air. So it reaches the end and stops, since it is only driven by the potential difference $$F=\frac{dU}{dx}$$ in it's rush to ...

3

So, we have series LCR circuit. $V$ is a constant voltage source. $L$, $C$, and $R$ represents the inductance, capacitance and resistance in the circuit respectively. A current $I$ flows through the circuit. Now, the current through each component is the same. So, the potential difference between each component added up together gives the emf $V$. ...

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From Kirchhoff's second law, the sum of all the voltages around a loop is equal to zero. That is, the sum of the voltages across the three elements of your circuit, R, L and C, must be equal to the time varying voltage from the source: $$V_R+V_L+V_C = V(t)$$ As $V_R=RI$, $V_L=L\frac{dI}{dt}$ and $V_C=\frac{Q}{C}$, we get your equation, which is correct: ...

3

The fact that the cell has internal resistance & it acts as a source corrupts your argument. Your argument that connecting a resistance across one of the cells in parallel reduces the overall resistance of the circuit and hence expect the current to increase is wrong. I will derive an equation to obtain the current & potential drop across the bulb ...

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But I wanted to know whether we can charge a capacitor while it is in use If, by "while it is in use", you mean while the capacitor is discharging, i.e., energy is flowing out of the capacitor to some load, then the answer is no since, by definition, if a capacitor is charging, energy is flowing into the capacitor. Put another way, a capacitor cannot ...

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The best way to solve these problems is to use the rules of resistors in series or parallel, and reduce it down into much simpler systems. When we have resistors in parallel, we can use the rule $$\frac{1}{R_{Total}} = \sum_i \frac{1}{R_i}$$ And when we have the resistors in series, we can use the rule $$R_{Total} = \sum_i R_i$$ We cannot make one ...

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Try redrawing it as so - should be easier to figure out. EDIT the top R2 in the second diagram should be R4.

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She tried touching the machine in various places, again nothing. I inadvertently touched her hand while she was touching the machine and then suddenly she felt it too. From this it is evident you were a good conductor to the ground. You later say : We came back out 15 minutes later after drinking our hot chocolate and tried to reproduce the ...

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Salt water is an excellent conductor. If you were still damp, and if the machine were broken in the right way, the salt water on your skin and slippers may have provided a route for a charge to flow to ground. Your wife with dry shoes, and you after drying out in the shop wouldn't.

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In the context of ideal circuit theory, the voltage across each resistor is given by voltage division: $$V_{R1} = 10V \cdot \frac{R_1}{R_1 + R_2}$$ $$V_{R2} = 10V \cdot \frac{R_2}{R_1 + R_2}$$ If we set $R_2 = \alpha R_1$, these equations become $$V_{R1} = 10V \cdot \frac{1}{1 + \alpha}$$ $$V_{R2} = 10V \cdot \frac{\alpha}{1 + \alpha}$$ which is valid ...

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I would try a more "dynamic" approach, where you actually have an RC circuit (https://en.wikipedia.org/wiki/RC_circuit ) - even a piece of wire has some non-zero capacitance. The typical time scale of transient processes in such circuits is $RC$. When $R\rightarrow \infty$, you have $RC\rightarrow \infty$, so you have to wait longer and longer for transients ...

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As R reaches infinity you will find new voltages appearing between R1 and R2, the result of induced interference from radio broadcast and other like environmental noise(s). As that soon to be free floating wire is more and more isolated it's less held back by the rest of the circuit and starts to behave like the free wire section it will soon become. ...

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If you send $R$ to infinity that's equivalent to an open circuit. Since an open circuit, as you can tell from the name, is not a closed loop, Kirchoff's law is not valid, hence your "paradox". Edit: Kirchoff's law is valid, but not in the form you wrote it. You cannot say that $V_R=5 V$ because when you say that $R \to \infty$ what you are saying is that it ...

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I think the question is quite interesting actually. When resistances are equal the voltage will divide itself equally because the resistances are coupled in series. Then we take each R to infinity and we assume that we do this to each resistance in the same manner. I think that the voltage across each resistance remains 5 Volts in this case when you take R ...

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The information you are given was actually vague at a mathematical level, but with your resources you have you already have a mostly correct grasp of the physical quantities. The most important variables in a circuit are voltage ($V$) and current ($I$), because you can measure them unlike the charge. As you already stated $I = \frac{Q}{t}$, and its a ...

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$Q=CV \Rightarrow \frac {dQ}{dt} = I = C \frac {dV}{dt}$ The voltage across the capacitor is equal to the voltage of the supply. So whatever the voltage of the supply does the voltage across the capacitor exactly follows. At time $t_A$ the capacitor is uncharged and the voltage across the capacitor is zero. However at this time the voltage of the supply ...

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Alternating circuit voltage and current periodically reach their respective maximum values after a certain time interval. Consider the current I=I(m)sin(wt) and Voltage V=V(m)sin(wt+pi/2)= v(m)cos(wt). What this phase difference signifies is that if you start the circuit at t=0, then the voltage at t=0 v=V(m) while i=0, meaning the circuit will gain maximum ...

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Suppose the emf of a cell is 1.500 V and its internal resistance is 10 $\Omega$. Connecting a voltmeter of resistance 10 M$\Omega$ across the cell will result in a voltage drop across the internal resistance of the cell of approximately $1.5 \times 10^{-6}$ volt which is hardly going to affect the reading on the voltmeter. On the other hand if the resistance ...

2

I took the liberty of redrawing this circuit for clarity. Can you take it from here?

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$R_1=1.5 \Omega$ , $R_2=1.5\Omega$ and $R_3=1.5\Omega$ gives you $$R_{eq} = \frac1{\frac1{1.5}+\frac1{1.5}+\frac1{1.5}}=\frac12=0.5 \Omega$$ You have to use this formula since the different resistors are parallel circuits.

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$$R_{eq} =\frac R3$$ $$R = 3R_{eq}$$ when $$R_1=R_2=R_3=R$$

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Remembering that resistance = $\frac V I$ work the resistance at $I=1$ and $I=2$. For the resistance to be constant the current-voltage characteristic must be a straight line and go through the origin. There is another parameter which is useful in some instances and that is called the incremental resistance $\frac {\Delta V} {\Delta I}$ which is related ...

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The 6V in 6V battery is a label which gives an indication of the sort of voltage which might be obtained from such a battery. For example if your ^ V battery was an lead-acid battery and it was fairly new and fully charged its voltage would be 6.3 volt and if older or partially discharges then it is more likely to be 6 V. A 1.5 V alkaline battery at the ...

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$n$ and $e$ are properties of the material and they do not change. Suppose you do halve the area to $\frac A 2$. Think of the situation before when it was area $A$ equivalent two conductors in parallel each of area $\frac A 2$ and each part carries a current $\frac I 2$ Now what is it about the conductor with area $\frac A 2$ which makes you think it has ...

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The resistance of a given wire is more than the rate of collisions alone. It also depends on the wire cross section. If you have twice as many electrons drifting you have twice the current for a given potential over the wire. In other words the resistance is half. You might be confused with conductivity or resistivity which is an intrinsic material ...

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The formulas are of course all true if used and interpreted correctly. But human error is the wild card. For practical reasons, I-squared-R is the most reliable formula because it's almost impossible to apply it incorrectly.

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After solving problems on circuits and power dissipation in them, I observed that $V^2/R$ is used when the voltage is constant across the elements in the circuit and $I^2R$ is used when current is constant through the elements in the circuit. They yield the same result when a purely resistive load is used. Even the formula $P=VI$ will give the answer. This ...

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When the switch is closed the two resistors on the left are in parallel. The current from the battery will split up half passing through one resistor and half passing through the other. The result will be a smaller voltage drop across each resistor, think $V=IR$. The measured voltage will less than it was with the switch open but not half as much ...

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EMF or Terminal voltage will be considered same if battery has no internal resistance. If battery has some internal resistance then terminal voltage will be different(less) from the EMF or potential difference from the battery. If a battery has internal resistance then what will be considered if the same statement is given.

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In your questions all the cases are assumed to be ideal unless mentioned. Therefore the electromotive force and the terminal voltage are equal in that case as internal resistance of the battery is considered negligible(if not given) . If the battery has internal resistance then the emf remains constant,but the terminal voltage decrease by a value which is ...

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in series the current is the same through each resistor. Not just the same, the current through each is identical. So that the lightbulbs will all have the same brightness but dimmer than if there was just one bulb on there. Well of course, series connected resistances add and so, the total resistance of two series connected bulbs is greater ...

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Light bulbs, or any loads, in series will all have the same current. This is unrelated to Ohm's Law - it's Kirchhoff's Current Law and it applies if the loads are ohmic or not. Assuming your source voltage stays the same, adding bulbs in series will increase the total resistance which will decrease the total current and make all the bulbs dimmer. The ...

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I can't understand at all everything after "As the resistance increases ...". Is that really how it was explained? Nonetheless there's an interesting point here. The analysis is not exactly the same as for an ohmic resistor, but not for the reason you suggest. The thing to consider is that the resistance of the light bulb depends on the temperature ...

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You are correct that light bulbs are non-ohmic (they don't obey Ohm's Law). But that makes no difference. The same current flows through each, even if they have completely different resistances. Electrical current (charge per second) is like the flow of a river. If there are no leaks, and no tributaries joining the river, then the volume of water per ...

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