Chapter 11 Electricity Questions-04
QUESTIONS
1. Draw a schematic diagram of a circuit consisting of a battery of three cells of $2 V$ each, a $5 \Omega$ resistor, an $8 \Omega$ resistor, and a $12 \Omega$ resistor, and a plug key, all connected in series.
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Answer
Three cells of potential $2 V$, each connected in series therefore the potential difference of the battery will be $2 V+2 V+2 V=6 V$. The following circuit diagram shows three resistors of resistances $5 \Omega, 8 \Omega$ and $12 \Omega$ respectively connected in series and a battery of potential $6 V$ and a plug key which is closed means the current is flowing in the circuit.
2. Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the $12 \Omega$ resistor. What would be the readings in the ammeter and the voltmeter?
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Answer
An ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the resistor it should be connected in parallel, as shown in the following figure.
The resistances are connected in series.
Ohm’s law can be used to obtain the readings of ammeter and voltmeter. According to Ohm’s law,
$V=I R$,
Where,
Potential difference, $V=6 V$
Current flowing through the circuit/resistors $=I$
Resistance of the circuit, $R=5+8+12=25 \Omega$
$I=V / R=6 / 25=0.24 A$
Potential difference across $12 \Omega$ resistor $=V_1$
Current flowing through the $12 \Omega$ resistor, $I=0.24 A$
Therefore, using Ohm’s law, we obtain
$V_1=I R=0.24 \times 12=2.88 V$
Therefore, the reading of the ammeter will be $0.24 A$.
The reading of the voltmeter will be $2.88 V$.
