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You're right: there's more to it than charge conservation. Consider a water tank with pipes connected to it below the water level. Despite conservation of mass, the sum of mass flow rates for water en …

It's fine to use 𝐸⃗ = 𝜌𝑗⃗ for a non-ohmic, isotropic material, as long as you realise that $\rho$ is, in this case, not a constant, but depends on the magnitude, $E$, of the field strength. I don't …

A voltmeter connected between a and b (see diagram) would indicate a potential difference between these points, that is a loss of electrical potential energy per unit (positive) charge going from a to …

The electric field depends on the potential difference (voltage) between the wires in the cable, not on the current going through the wires. The electric field will be strongest inside the cable, betw …

Let $N_1\Phi_{11}$ be the flux linkage with coil 1 due to current in coil 1, and $N_1\Phi_{12}$ be the flux linkage with coil 1 due to current in coil 2 and so on. Therefore, $$N_1\Phi_{11}=L_1I_1\ \ …

The emf, $\mathscr E$, of an electrical supply is the gross energy transferred per unit charge flowing through the supply from some other form (e.g chemical in a battery, mechanical work for a generat …

You've done all that is needed in your first paragraph – essentially you have treated the top and bottom rows of resistors as potential dividers. In my opinion the Wheatstone bridge approach adds noth …

For temperatures up to a few hundred kelvin, the resistance of a metal wire is roughly proportional to its kelvin temperature, $T$, so $$R=\alpha T,\ \ \ \ \ \text{that is}\ \ \ \ \ \tfrac VI=\alpha T …

Ohm's law in the form $V=IR$ can be regarded as a special case of the arguably more fundamental local relationship $\vec J = \sigma \vec E$ in which $\vec J$ is the current density, $\sigma$ is the co …

"If a source provides some energy, and all of it is dissipated as heat, is it of any use?" Do you not consider electric heaters, (eg fan heaters, convectors) useful? They consist, essentially, of a le …

The current through the inductor, L, at any instant is the same as that in the wires of the capacitor, C, as the L and the C are parts of a series circuit. If that current is $I=I_0 \cos \omega t$, th …

Usually the output from each stage of the amplifier is a constant voltage (with respect to some reference point, which I'll call 'the ground rail') with an alternating 'signal' voltage superimposed on …

"Am I supposed to find the voltage across each resistor first?" Yes, that's a good first step. [The quick way is to treat the left and right resistor combinations as potential dividers.] Note that the …

I apologise for not reading your question properly: this answer was first written for an inductor connected across an ideal dc supply. Nevertheless it deals with the paradox you have presented. You ar …

(a) "Is this correct?" Yes, if we assume that each bulb has a fixed resistance. But note that the voltage across the third bulb (the one in series with the parallel combination) is twice what it was w …