8.1 Three identical light bulbs, A, B and C, are each rated at 6 W, 12 V - NSC Physical Sciences - Question 8 - 2019 - Paper 1

To find the resistance of each bulb, we use the formula:

$P = \frac{V^2}{R}$

Rearranging this gives:

$R = \frac{V^2}{P}$

Here, the voltage $V = 12 V$ and power $P = 6 W$, thus:

$R = \frac{12^2}{6} = \frac{144}{6} = 24 \Omega$

So, the resistance of each bulb is 24 Ω.

Using the total resistance in the circuit, we have:

$R_{total} = R_{b1} + R_{b2} + R_{b3} + r$

where each bulb has a resistance of $R = 24 \Omega$ and the internal resistance $r = 2 \Omega$.

Therefore:

$R_{total} = 24 + 24 + 24 + 2 = 74 \Omega$

Now apply Ohm's law to find the total current, $I$:

$I = \frac{V}{R_{total}} = \frac{12}{74} \approx 0.162 A$

The voltage across each bulb in series is given by:

$V_C = I * R$

Using the current we just calculated, $I \approx 0.162 A$, and the resistance $R = 24 \Omega$:

$V_C = 0.162 * 24 \approx 3.89 V$

Light bulb C will not burn at maximum brightness because the potential difference across it is less than the rated voltage of 12 V. As the current flowing through C is reduced, the brightness diminishes since brightness is directly proportional to both current and voltage.

The current in resistor A is greater than that in resistor C because they are in parallel branches and tend to split the total current according to their resistances. Since A has a lower resistance compared to C, more current will flow through A.

If resistor C is removed, resistor B will experience an increase in current. The current flowing through A was split due to the presence of C; without C, more total current will flow through A and hence more will also flow through B since they are connected in parallel.