Draw a circuit diagram for a circuit consisting of a battery of five cells of $2$ volts each, a $5\, \Omega,$ resistor, a $10\, \Omega$ resistor and a $15 \,\Omega$ resistor, an ammeter and a plug key; all connected in series. Also, connect a voltmeter to record the potential difference across the $15\, \Omega$ resistor and calculate

$(i)$ the electric current passing through the above circuit and

$(ii)$ potential difference across $5\, \Omega$ resistor when the key is closed.

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The diagram is as shown

The equivalent resistance of the series combination of $R _{1}, R _{2}$ and $R _{3}$ is

$R = R _{1}+ R _{2}+ R _{3}$

$=5+10+15=30$ $ohm.$

Therefore, current through the circuit is

$I=V / R=10 / 30=0.33 A$

Now, potential difference across $5 \Omega$ resistor is

$V=I R_{1}=0.33 \times 5=1.67 V$

1091-s224

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