State Ohm's law - Leaving Cert Physics - Question c - 2007

In a series circuit, the total resistance (R_total) is the sum of the individual resistances. Thus, for the resistors of 3 Ω and 9 Ω:

$R_{total} = R_1 + R_2 = 3 \, \Omega + 9 \, \Omega = 12 \, \Omega.$

Therefore, the total resistance of the circuit is 12 Ω.

Using Ohm's law, the current (I) can be calculated by rearranging the formula:

$I = \frac{V}{R}$

Substituting the values:

$I = \frac{6 \, V}{12 \, \Omega} = 0.5 \, A$

Thus, the current in the circuit is 0.5 A.

The potential difference (V_R) across the 9 Ω resistor can be calculated using Ohm's law:

$V_R = I imes R = 0.5 \, A \times 9 \, \Omega = 4.5 \, V$

Thus, the potential difference across the 9 Ω resistor is 4.5 V.

An instrument used to measure potential difference is a voltmeter or a multimeter.