ICSE Class 10 Physics 2016 Paper

A battery of emf $12\ \mathrm{V}$ and internal resistance $2\ \Omega$ is connected with two resistors $A$ and $B$ of resistance $4\ \Omega$ and $6\ \Omega$ respectively joined in series.

Find:

Question : 62 of 73

Marks: +1, -0

The terminal voltage of the cell.

Solution:

\;\; \text{ Voltage drop Ir } \; = 1 \times 2 = 2 \; \text{ volt } \;

\; \text{ Terminal Voltage } \;\; = \; \text{ emf } \; - \; \text{ Voltage drop } \;

\; = 12 - 2 = 10 \; \text{ volts. } \;

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