Answer to Question #107391 in Electric Circuits for Jennifer

Question

A cell of e.m.f E and internal resistance, r, was connected in series with two external resistors X and Y of 2Ω and 3Ω respectively. A high - resistance voltmeter connected across Y reads 1.0 V. When another resistor of 2Ω was connected parallel to X the voltmeter read 0.4 V. Calculate the e.m.f and internal resistance of the cell

Answer 1

The total resistance of the circuit is $r+2+3=r+5$

current in the circuit =$\frac{E}{r+5}$

we are given that the drop across 2 ohm resistor is 1 ohm

hence $\frac{E}{r+5}\times2=1$

$2E=r+5\\2E-r=5$ .............(1)

in the next case a 2 ohm resistor is connected in parallel to the the resistor X(2ohm ), hence the equivalent resistor across X is equal to 1 and the total resistance across the circuit is equal to $r+1+3=r+4$

current in the circuit = $\frac{E}{r+4}$

Since a 2 ohm resistor is in parallel to the 2 ohm(X) resistor hence the current across X will get halved

current across X = $\frac{E}{2(r+4)}$

and now the drop across X is 0.4 volts

hence $\frac{E}{2(r+4)}\times2=0.4$

$E=0.4\times(r+4)\\E-0.4r=1.6$ .............(2)

from 1 and 2 we get

E=2volts and r=1ohm