Current $l$ versus time $t$ graph through a conductor is shown in the figure. Average current through the conductor in the interval $0$ to $15 \,s$ is ............ $A$
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(d)
$\Delta q=$ Area $( l / t )$
$\Delta q=\frac{1}{2} \cdot 10 \times 15=75 \,C$
$i_{\text {avg }}=\frac{75}{15}=5 \,A$
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