Tag Info

New answers tagged

2

How do I identify which ones are parallel or series? If all of the current leaving one resistor enters another resistor, the two resistors are in series. The resistances of series connected resistors can be added together to find the equivalent resistance of a single resistor, e.g., $$R_{eq} = R_1 + R_2 $$ If all of the voltage across one resistor is ...


0

Since this is a homework type question, I can only offer you a tip and not solve it for you. Tip - Look at this series combination for example Notice that there is some current $I$ flowing through each of these three, and the value of potential at points 1, 2, 3 and 4 (across $R_1$, $R_2$ and $R_3$) aren't the same. On the other hand, in a parallel ...


0

You want to work this out in stages. Start at the right hand side where there are three resistors clearly in series. "In series" means that the current that flows through one resistor is equal to the current flowing through the other resistor. The simplest way to identify this is to look for nodes (points where components are connected) with just two wires: ...


0

We know that the tension between a perfect current source is 0 That's a common misconception held by those first introduced to current sources. In fact, an ideal current source produces a fixed current regardless of the voltage across. Put another way, the voltage across an ideal current source is determined by the attached circuit which is, in this ...


16

It depends on the internal resistance of the source. Fist consider a "voltage supply". What does "voltage supply" even mean? A voltage supply is supposed to output a fixed voltage no matter what we connect it to. Is this even possible? Suppose we connect the two terminals of the voltage supply together through a piece of wire, i.e. a really low resistance ...


1

You are right, as is Eoin's answer. I'm only answering to show one way to think about it that is useful to me when people bring up this common misconception. Imagine you have a pair of terminals with some voltage between them (like a common 1.5V battery). Nothing's connecting those two terminals together. Except for air, that is. Is there any current? No, ...


3

It depends on if your power supply is constant voltage or constant current. Usually, it's the former, so it means that the $P = V^2/R$ is the more appropriate one to use. Therefore, a smaller resistor will dissipate more power in this situation. In some situations (electromagnets is one that comes to mind), the load is driven with constant current, so ...


0

What is the easiest way to solve this? The easiest way is unlikely to be the most illuminating way. What I propose is that you first work the problem from basic circuit principles and then explore more powerful solution methods such as, e.g., node voltage analysis, Thevenin equivalent circuit etc. Start from the right hand side with Ohm's law: $$I_6 ...


1

You were given the current through $R_1$ as 1mA. You should know that the 1mA current is shared between $R_2$ and $R_3$. You could determine the voltage drop across $R_1$, and then $R_{eq}$ (the equivalence of $R_2$ and $R_3$) so you can calculate the current through each resistor. Added: Yes, the voltage drop across $R_1$ is 6V. Remember that this circuit ...


1

What is the smallest resistance that can be used in a 5 V circuit if the resistors available are rated for 1/4 W? This is a trick question of sorts. It's true that one could work from the relationship $$p_R = \frac{v^2_R}{R}$$ and the range of $R$ for which $p_R \le 0.25W$. However, this completely ignores the issue of derating for reliability. ...


1

Why do we use this formula to find the total resistance? Because parallel connected conductances sum just as series connected resistances. Ohm's law is $$V_R = RI_R $$ The dual of Ohm's law is $$I_G = GV_G $$ Since, for parallel connected circuit elements, the voltage across is identical, it follows that, for parallel connected conductances ...


1

Think about current flow. If we take each individual resistor and determine the current for the applied voltage, we get: $$I_T=\frac {V}{R_1} +\frac {V}{R_2} + ...$$ Dividing everything by the voltage give us: $$\frac {I_T}{V}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Which is the same as: $$\frac {1}{R_{eq}}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ We now need ...


1

This is derived from the the formula for calculating total resistance in a parallel circuit $$\frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$ taking its inverse gives: $$R_t = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}}$$


0

You have the right equation - now just solve for $R$. This means you multiply both sides by $R$, and divide both sides by $P$. In steps, you started with $$P = \frac{V^2}{R}$$ Multiply both sides by $R$: $$P\ R = \frac{V^2}{R}R = V^2$$ Now divide both sides by $P$, to get $$\frac{P}{P}R = \frac{V^2}{P}\\ R = \frac{V^2}{P}$$ Physics equations manipulate ...


1

Yes I also think that's what they did. And yes, you'd also change the voltage proportionally to the current. But all of this is as long as the resistance is constant (which is what they mean when they say the circuit is the same). A less-prone-to-mistake equation to use in these cases where the circuit is kept the same is gotten by substituting Ohms law, $V ...


0

Idealizations are just that, approximately true in many situations. If you connect a wire across a battery, the non-ideal nature of the wire and battery will show-the wire will have some resistance and the battery will not be a perfect voltage source. In most circuits, the resistance of the other circuit elements are much higher than the resistance of the ...


1

What you are missing is called conservation of charge. The fundamental fact is that electrons do not simply 'disappear' or vanish into thin air inside a resistor. This means that, if there is a certain number of electrons flowing into the resistor, these electrons must also flow out on the other side. In particular, in a steady state (i.e. the flow of ...


1

An ohm is one volt per amp. That is, a one ohm (ideal, linear) resistor requires 1 V in order for 1 A to pass through it. A 2 $\Omega$ resistor requires 2 V in order for 1 A to flow. Conversely, if 1 A is forced to flow through a 1 $\Omega$ resistor, 1 V will be developed across the resistor. Does resistance decrease the amount of electrons passing ...


2

An intuitive tool to understand resistivity in the classical context, is Drude model. Consider a 3 Dimensional lattice, of stationary obstacles, that are separated by a characteristic length $\lambda$. Now consider a charged particle, operating under a potential difference that creates a constant E-field in this 3D lattice, let's say in the $\hat{x}$ ...


0

When the current is turned on for 10 seconds the capacitors have gradually charged according to the equation Q(t)=CV(1-e^(-RC)) Now after 10 seconds when the switch is open the two capacitors act as two voltage sources in parallel and the current could be found out by superposition principle which is used when we have more than one source


1

After the switch is opened, we have a series circuit - the current through each circuit element is identical and equal to $i(t)$. Choosing a clockwise reference direction for the series current, KVL clockwise yields $$\frac{1}{1\mu F}\int_{-\infty}^t i(\tau) d\tau + i(t) \cdot 1k\Omega + i(t)\cdot10k\Omega + \frac{1}{1000\mu F}\int_{-\infty}^t i(\tau) ...


1

Here the capacitors look like in series but they are not actually. Capacitors can be said to be in series only if they carry same amount of charge which is not the case here.If you calculate charges will come out to be different.Look at the time constants(R*C) for the first branch and second branch, they are 1ms and 10sec respectively. As switch is on ...


5

When the switch is opened, the circuit is the equivalent of this, so I think you can clearly see that the resistors are in series and so are the capacitors. It seems that you already understand how to calculate series resistance, so I'll show how to understand series capacitance. First, you probably already know that capacitance is defined as the ratio ...


0

The best way to go about these kind of problems, is to use Kirchoff's loops. Whenever there is ambiguity about parallel/serial, Kirchoff is the way, since you disregard the question of serial vs. parallel altogether. Also remember that while resistors observe: $$ \frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2} $$ for parallel and, $$ R_T=R_1+R_2 $$ for serial, ...


0

If you cannot see that this is a Wheatstone bridge (2:4 = 3:6, so the two "arms" are balanced and there is no current in the 5 Ohm resistor) then you can use a simple analysis with just two unknowns: the voltage at the two nodes between A and B, call them x (where 2,4,5 meet) and y (3,5,6 meet). Given an imposed voltage V at A, and 0 (ground) at B, you can ...


0

look at the final equivalent circuit and solve you will get c)39/7


4

The answer is simple, the resistor doesn't know what voltage drop to provide, and to some extent it doesn't "care" either (unless it's so high that the current fries it but that's another issue). The "voltage drop" is only governed by the potential difference created by the generator (battery or other). If the circuit is open, the electrons have no path so ...


3

It's not a dumb question. The electrons at the negative side of the battery have energy, and will look for "any path" to get rid of that energy - move in the direction of an electric field. When you create a circuit, you move the potential of the positive terminal closer to the negative terminal. If you assume that "wire" has no resistance (for simplicity) ...


3

The individual resistances are all positive, so the sum $$ \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots \,,$$ is larger than the inverse of any of the individual resistances, and that means that the inverse of the sum is necessarily smaller than any of the resistances. No mucking around with the two-resistor form required.


3

Think about current flow. If we take each individual resistor and determine the current for the applied voltage, we get: $$I_T=\frac {V}{R_1} +\frac {V}{R_2} + ...$$ Dividing everything by the voltage give us: $$\frac {I_T}{V}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Which is the same as: $$\frac {1}{R_{eq}}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Since there ...


2

We can prove it by induction. Let $$ \frac{1}{R^{(n)}_{eq}} = \frac{1}{R_1} + \cdots+ \frac{1}{R_n} $$ Now, when $n=2$, we find $$ \frac{1}{R^{(2)}_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \implies R_{eq}^{(2)} = \frac{R_1 R_2}{R_1+R_2} = \frac{R_1}{1+\frac{R_1}{R_2}} = \frac{R_2}{1+\frac{R_2}{R_1}} $$ Since $\frac{R_1}{R_2} > 1$, we see that $R^{(2)}_{eq} ...


3

This is solved using start-delta transformation and Delta-Star transformation like shown in the below picture when you rewrite the circuit diagram it looks like this and you can notice delta network and star network in it. Applying proper transformation reduces the complex network to simple network, then you can solve simple series/parallel combination ...


1

If $E$ is the emf (open-circuit voltage) of the battery and $r_s$ is the internal resistance, the equation relating the series current $I_S$ through an external resistance $R_L$ is given by: $$I_S = \frac{E}{r_s + R_L}$$ Now, the voltage across the battery terminals $V_{BAT}$ is given by $$V_{BAT} = E - I_S \cdot r_s = E\frac{R_L}{r_s + R_L}$$ So, even ...


0

When current flows, inside the source there is always some voltage drop, so the voltage on the terminals is lower than in the static situation. We model this behaviour by simple model depicted on your picture; we assume that the real source consists of idealized source -| |- that gives the emf and has not resistance, and of ordinary ohmic resistance (2 ...



Top 50 recent answers are included