First slide

ammeter

Question

A galvanometer together with an unknown resistance in series is connected across two identical batteries each of 1.5 V. When the batteries are connected in series, the galvanometer records a current of 1A, and when the batteries are in parallel, the current is 0.6 A. What is the internal resistance of the battery ?

Moderate

A

$1 Ω$
B

$\frac{1}{2} Ω$
C

$\frac{2}{3} Ω$
D

$\frac{1}{3} Ω$

Solution

Is $\begin{array}{l} = \frac{2 E}{2 r + R} = 1 A \\ = \frac{3}{2 r + R} = 1 \end{array}$
$I_{p} = \frac{E}{\frac{r}{2} + R} = 0.6 A = \frac{3}{r + 2 R} = 0.6 A$
$S o l v i n g r = 1 / 3 Ω$

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