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The internal resistance of the cell, E in Fig. 3 is 0.5 ohms.

The internal resistance of the cell, E in Fig. 3 is 0.5 ohms.
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Determine the ammeter reading when the switch, S is closed.

Answers


John
E = IR + Ir

I = $\frac{E}{R+r}$ = $\frac{2.0}{2.0 \times 0.5}$ = 0.8A
johnmulu answered the question on May 23, 2017 at 07:17

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