Question

Three cells each of e.m.f $1.5V$ are first connected in series and then in parallel to an external resistance. The currents obtained in the two cases are respectively $1A$ and $0.36A$. The external resistance is:

• A. 1.25$\Omega$
• B. 3.26$\Omega$
• C. 4.125$\Omega$
• D. 2.235$\Omega$

Answer 1

Answer: (C)

4.125$\Omega$

Let the internal resistance of each cell be $r$

In case of series connection

$\Rightarrow 3E=I[R+3r]$

$I_1=1A$

$\Rightarrow 3E=R+3r$

$=4.5=R+3r$ ------- A

Applying kvl for outer loop

3IR + Ir $=$ E

I[3R+r] $=$ E

0.12[3R+v] $=$ E

3R + r $=\frac{50}{4}=12.5$

3R + r $=$ 12.5 - B

Solving A & B

R+3 [12.5-3R] $=$ 4.5

$R+37.5-9R$ $=$ $4.5$

$8R$=$33$

$R=\frac{33}{8}=4.125\Omega$