Question

The potential difference applied across a given resistor is altered so that the heat produced per second increases by a factor of 9. By what factor does the applied potential difference change?
(b) In the figure shown, an ammeter A and a resistor of 4$\Omega$ are connected to the terminals of the source. The emf of the source is 12 V having the internal resistance of 2 $\Omega$
Calculate the voltmeter and ammeter readings

Answer 1

(a) Heat produced is given by,
$H=\frac{V^2}{R}t\Rightarrow H\propto V^2\frac{H^{\prime }}{H}=\frac{V^{\prime 2}}{V^2}$
$\frac{9H}{H}=\frac{V^{\prime 2}}{V^2}[\because H^{\prime }=9H]V^{\prime 2}=9V^2{V}^{\prime }=3V$
So , potential difference is increased by a factor of 3.
(b) Current , I =$\frac{E}{R+r}$
I = $\frac{12}{4+2}=\frac{12}{6}=2A$
Ammeter reading = 2A
V = E - Ir
V = 12-2 $\times$ 2
= 12-4
V = 8 V
Volmeter reading = 8V.