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: