ICSE Class 10 Physics 2016 Solved Paper

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

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Question : 62 of 73

Marks: +1, -0

The terminal voltage of the cell.

Solution:

\;\;\text " Voltage drop Ir "\;=1 × 2=2 \;\text " volt "\;

\;\text " Terminal Voltage "\;\;=\;\text " emf "\;- \;\text " Voltage drop "\;

\;=12-2=10 \;\text " volts. "\;

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