# Tag Info

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take only first 2 resistors and rest as $x$ now the vertical resistor and your $x$ will be in parallel, effective resistance would be $Rx/R+x$ with series in horizontal resistor. Now equivalent resistance would be $$Req. = R+ (Rx/R+x).$$ take Req. as $x$ again Form quadratic equation and solve for $x$. This will be the answer.

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Correct me if im wrong The circuit can be reduced to Current should be I = V/R = 4/3V The voltage drop across the 6 resistor is 6/9 * 12 = 8 and the voltage drop over the 3 should be 12-8 = 4 Then we look at the circuit again The Voltage drop across each 'section' of parallel resistors is the same. So V across the 8 and 4 is 4. To calculate Vo ...

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The Lorentz force on a charge in an electromagnetic field is $$F=q(E+v \times B) \ \ .$$ For an electron between the cylinders, $q$ is negative, and $E$ is defined as pointing outward, so the electron will experience a force radially inward. But due to the unfortunate sign convention used for currents, electrons flowing inward means that the conventional ...

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Yes - electrons flow from the negative to the positive, so in the opposite direction to the conventional direction of the electric field (which points from positive to negative). So if the E field points outwards, the electrons flow from the outer to the inner cylinder. The direction does not affect the answer (the calculation of the flow) though - at least ...

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There are two basic systematic approaches for analyzing a circuit like this: nodal analysis, which uses Kirchhoff's current law (KCL), and mesh analysis, which uses Kirchhoff's voltage law (KVL). The two methods would work about equally well for this particular circuit, in that both methods would require solving a system of three equations in three ...

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The admittance looking into the network "$Y_{net}$" can be expresed as: $$Y_{net} = \dfrac{1}{R + \dfrac{1}{Y_{net} + Y_{diode}}}$$ Rearranging as a quadratic and solving: $$Y_{net}^2R + Y_{net}Y_{diode} - Y_{diode} = 0$$ $$Y_{net} = \dfrac{\sqrt{Y_{diode}}\sqrt{4R +Y_{diode}}}{2R}$$ Where $$Y_{diode} = \frac{nV_t}{I_s}e^{-\frac{V_f - I_fR}{V_t}}$$ ...

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This is because what changes in each resistor is the current passing through and not the voltage difference. One the other hand when resistors are in series they have the same current passing through, but different voltage through each one's nodes. In essence when resistors are in parallel do not share same current path (i.e wire) but share same voltage. On ...

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Kirchoff's laws tell us that the potential drop across any closed loop in a circuit must be equal to the voltage sources in the loop, from which we conclude that the voltage drop across resistors in parallel must be equal. Ohm's law states: $$V=IR$$ From which we conclude that, since $V$ is fixed, if the different resistors have different $R$'s, then the ...

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As radio amateurs we've all learned the various relationships of power, voltage, current and resistance as expressed in Ohm's Law Ohm's law is: $$E = IR \tag{1}$$ This doesn't directly say anything about power. There is the related Joule's first law, which relates to electrical power converted to heat in resistive materials: $$P = I^2 R \tag{2}$$ ...

3

As Kevin Reid aptly explains, the circuit you have drawn is not realizable. But, let's take the closest physical thing you could build, assuming: your voltage source can supply enough energy that we don't hit its limits like all physical things, this apparatus has non-zero size Then, the circuit you actually built is this: simulate this circuit ...

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… an ideal power source capable of providing infinite current with no drop in the voltage it supplies. … Let's ignore the effects of current density on superconductors for now. … In these phrases is the explanation for the contradictory possibilities you have computed: you have supposed an impossible circuit. As a mathematical model, the behavior of ...

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“[A]n equal and opposite field” is an oxymoron and gibberish. Henceforth I’ll assume that we have a chemical voltage source there. The Coulomb’s law is applicable in the case of a charged body, a body that has a constant electric charge. A chemical voltage source is a completely different situation: there is no pre-defined distribution of charges (as they ...

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The measure $R$ of resistance is an invented one. It was deduced long ago by experiment that many materials had a constant ratio $\frac {V}{I}$ between the voltage applied and the current flowing. Thus the quantity 'resistance' was defined to be precisely this ratio. Later, when inductive and capacitive effects were observed, 'reactance' $Z$ was defined ...

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$P = IV$ applies to all circuit branches. $P = I^2R$ or $P = V^2/R$ are restatements of the general rule that apply when we are considering power delivered to an ideal resistor that behaves according to Ohm's law $V = IR.$ I have seen in some circuit $V^2/R$ is not equal to $I^2R$ (like when there is capacitor or inductor). Why is that? Those ...

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What is the flaw in my thinking? The voltage across the capacitor in the series RC circuit given, assuming zero initial capacitor voltage, is given by $$v(t) = E\left(1 - e^{-\frac{t}{RC}} \right), t \ge 0$$ Note that $v(t) \rightarrow E$ as $t \rightarrow \infty$. The energy stored in the capacitor, as a function of time, is $$U(t) = ... 1 Once the capacitor has fully charged the current in the circuit will be zero, so the voltage drop across the resistor is zero and hence the voltage across the capacitor is equal to the cell voltage. Having said this, the current falls exponentially with time so in principle the current takes an infinite time to fall to zero, and the voltage across the ... 0 For a constant resistance Ohm's Law is$$V = IR. Now, it happens that it's pretty easy to make a constant-resistance device (we call them "resistors") and that it's easier to make constant-voltage devices than constant-current devices. So most of the circuit problems we encounter have a constant-voltage device like "a wall socket" or "a battery" or "a ...

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Ohm's law can be summarized as this: The drop in electric potential (voltage) across a section of a circuit is equal to the total current flowing through that section of the circuit multiplied by the total resistance of the section of the circuit. By that law, if you have a current flowing and there is some net resistance, there will be a drop in voltage. ...

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Increasing the number of turns increases the magnetic field if the current remains constant. In your situation, you are postulating (implicitly) that the applied voltage is constant, and that the current is reduced. The product $N\cdot I$ is therefore unchanged, and the magnetic field does not increase when you increase the number of turns of a resistive ...

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