Chapter 11 Electricity Questions-04

QUESTIONS

1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2V each, a 5Ω resistor, an 8Ω resistor, and a 12Ω resistor, and a plug key, all connected in series.

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Answer

Three cells of potential 2V, each connected in series therefore the potential difference of the battery will be 2V+2V+2V=6V. The following circuit diagram shows three resistors of resistances 5Ω,8Ω and 12Ω respectively connected in series and a battery of potential 6V 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 12Ω 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=IR,

Where,

Potential difference, V=6V

Current flowing through the circuit/resistors =I

Resistance of the circuit, R=5+8+12=25Ω

I=V/R=6/25=0.24A

Potential difference across 12Ω resistor =V1

Current flowing through the 12Ω resistor, I=0.24A

Therefore, using Ohm’s law, we obtain

V1=IR=0.24×12=2.88V

Therefore, the reading of the ammeter will be 0.24A.

The reading of the voltmeter will be 2.88V.



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