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# 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?

Net resistance in the circuit(R) = Sum of the resistances of all resistors
R= 5 $$\Omega$$+ 8 $$\Omega$$ + 12 $$\Omega$$ = 25$$\Omega$$

We have; $$R=\frac{V}I$$

$$25 \Omega =\frac{6V}I$$

or $$I=\frac{6V}{25 \Omega }=0.24A$$

Electric current remains the same through all resistors because resistances are connected in series.

Here we have,

Electric current, $$I$$ = 0.24A

Resistance, R = 12$$\Omega$$

So, potential difference, V through the resistor of 12 $$\Omega$$ = $$I$$ x R

Or, V = 0.24A x 12 $$\Omega$$ = 2.88 V

Thus, reading of ammeter = 0.24A

Reading of voltmeter through resistor of 12 $$\Omega$$ = 2.88V

And the circuit can be formed as: