Question

A cell of e.m.f. 2 V and internal resistance 1.2 Ω is connected to an ammeter of resistance 0.8 Ω and two resistors of 4.5 Ω and 9 Ω as shown in figure. Find:
a) The reading of the ammeter,
b) The potential difference across the terminals of the cell, and
c) The potential difference across the 4.5 ohm resistor.

Answer 1

Total resistance of the circuit:
The resistors of $4.5ohm\text{and}9ohm$
resistors are in parallel. Their equivalent
resistance $R_p$ ,
$R_p=\frac{4.5\times 9}{4.5+9}=3\Omega$
The internal resistance $r=1.2\Omega$, ammeter
resistance $R_A=0.8\Omega ,\text{and}Rp=3\Omega$ are in series.
Let $R_{eq}$ be the total resistance,
$R_{eq}=r+R_A+R_p=1.2+0.8+3=5\Omega$
Reading of ammeter:
Given,
emf of the cell $𝝴=2V$
Equivalent resistance of the circuit $R_{eq}=5\Omega$
Applying Ohm’s law for $R_{eq}=5\Omega$
$𝝴=I{R}_{eq}$
Current through the circuit
$I=\frac{𝝴}{R_{eq}}=\frac{2V}{5\Omega }=0.4A$
Reading of the ammeter is 0.4 A.
p.d. across the terminals of the cell:
The terminal voltage is the sum of the voltage across the battery ε and across the internal resistance r.
$V=\epsilon -Ir$
Given:
e.m.f. of the battery $\epsilon =2V$
Current (calculated) $I=0.4A$
Internal resistance $r=1.2$
So, the potential difference across the terminals of the battery,
$V=2V-(0.4\times 1.2)V=1.52V$
p.d. across the 4.5 ohm resistor:
The p.d. across the 4.5 ohm resistor will be same as across the parallel combination $R_p$ We know,
Current through 𝑅𝑝 is $I=0.4A$
Applying Ohm’s law for $R_p=3\Omega$
$V_{4.5\Omega }=Vp=IRp=0.4A\times 3\Omega =1.2V$