Draw a schematic diagram of an electric circuit. (in the "on" position) consisting of a battery of five cells of $2\, V$ each, a $5\, \Omega$ resistor, a $8\, \Omega$ resistor, a $12$ $\Omega$ resistor and a plug key, all connected in series. An ammeter is put in the circuit to measure the electric current through the resistors and a voltmeter is connected so as to measure the potential difference across the $12\, \Omega$ resistor.

Calculate the reading shown by the

$(a)$ ammeter $(b)$ voltmeter in the below electric circuit

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$(a)$ $R =5 \Omega+8 \Omega+12 \Omega=25 \Omega$

So $I=\frac{V}{R}=\frac{2 \times 5 V}{25 \Omega}=0.4 A$

$\therefore$ Ammeter reading $=0.4 A$

$(b)$ $V = IR =0.4 A \times 12 \Omega=4.8 V$

$\therefore$ Voltmeter reading $=4.8 V$

1091-s213

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