Question
  1. 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?
  2. 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 an internal resistance of $2\Omega$. Calculate the voltmeter and ammeter readings.

Answer

  1. $H=\frac{V^2}{R}$
$\therefore$ V increases by a factor of $\sqrt{9}=3$
  1. Ammeter Reading $I=\frac{V}{R+r}$
$=\frac{12}{4+2}A=2A$

Voltmeter Reading $V=E-Ir$

$=[12-(2\times2)]\text{ }\text{V}=8\text{V}$

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