ICSE Class 10 Physics 2019 Solved Paper

The diagram above shows a circuit with the key $k$ open. Calculate :

Question : 60 of 74

Marks: +1, -0

the current drawn from the cell when the key

k

is open.

Solution:

When the key

{\K}

is open,

I\;=\;{E}/{R+r}=\;{3.3}/{5+0.5}

\;=\;{3.3}/{5.5}=\;{3}/{5}=0.6 {\A}

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