Question

Two cells of emf 1.5 V each, having internal resistance $r$, when connected in series current through unknown resistance $R$ is 1 A and when they are connected in parallel current through 'R' is 0.6 A, then value of r is

• A. $\frac{1}{2} \Omega$
• B. $1 \Omega$
• C. $\frac{2}{3} \Omega$
• D. $\frac{4}{3} \Omega$

Answer 1

Answer: (A)

1/2 \Omega

Answer 2

In series connection, equivalent EMF $E_{eq,s} = 1.5 + 1.5 = 3$ V. Equivalent internal resistance $r_{eq,s} = r + r = 2r$. Current $I_1 = \frac{E_{eq,s}}{R + r_{eq,s}} \implies 1 = \frac{3}{R + 2r} \implies R + 2r = 3$ (1)

In parallel connection, equivalent EMF $E_{eq,p} = 1.5$ V. Equivalent internal resistance $r_{eq,p} = \frac{r \times r}{r + r} = \frac{r}{2}$. Current $I_2 = \frac{E_{eq,p}}{R + r_{eq,p}} \implies 0.6 = \frac{1.5}{R + r/2} \implies R + r/2 = \frac{1.5}{0.6} = 2.5$ (2)

Subtracting (2) from (1): $(R + 2r) - (R + r/2) = 3 - 2.5 \implies \frac{3r}{2} = 0.5 \implies 3r = 1 \implies r = \frac{1}{3} \Omega$.

Since $1/3 \Omega$ is not an exact option, we check the closest one. If $r = 1/2 \Omega$, then from (1), $R + 2(1/2) = 3 \implies R = 2 \Omega$. Check in (2): $R + r/2 = 2 + (1/2)/2 = 2 + 1/4 = 2.25 \Omega$. The expected value is $2.5 \Omega$. The current would be $1.5/2.25 = 2/3 \approx 0.667$ A, which is closer to $0.6$ A than other options.