(a) In this problem, we have a battery of 3 cells of 2V each, so the total potential difference (or voltage) of the battery will be `3 xx 2 = 6V`. The circuit consisting of a battery of three cells of 2V each (having a voltage of 6V), resistors of `5Omega, 8Omega, 12 Omega` and a plug key, all connected in series is given in Figure alongside.
(b) The above circuit can be redraw by including an ammeter in the main circuit and a voltmeter across the `12 Omega` resistor, as shown in Figure alongside. Please note that the ammeter has been put in series with the circuit but the voltmeter has been put in parallel with the `12 Omega` resistor. We will now calculate the current reading in the ammeter and potential difference reading in the voltmeter.
(i) Calculation of current flowing in the circuit. The three resistors of `5 Omega, 8 Omega` and `12 Omega` are connected in series. So,
Total resistance, `R = 5 +8 +12`
`= 25 Omega`
Potential difference, `V = 6V`
And, Current, `I = ?` (To be calculated)
Now, `(V)/(I) = R`
So, `(6)/(I) = 25`
`25 I = 6`
`I = (6)/(25)`
`I = 0.24A`
Now, since the current in the circuit is 0.24 amperes, therefore, the ammeter will show a reading of 0.24A.
(ii) Calculation of potential difference across `12 Omega` resistor. We have just calculated that a current of 0.24A flows in the circuit. The same current of 0.24A also flows through the `12 Omega` resistor which is connected in series. Now, for the `12 Omega` resistor:
Current, `I = 0.24A` (Calculated above)
Resistance, `R = 12 Omega` (Given)
And, Potential difference, `V - ?` (To be calculated)
We know that, `(V)/(I) = R`
So, `(V)/(0.24) = 12`
And `V = 0.24 xx 12`
`V = 2.88V`
Thus, the potential difference across the `12 Omega` resistor is 2.88 volts. So, the voltmeter will show a reading of 2.88 V.