8.1 How does the brightness of bulb P compare with that of bulb Q? Give a reason for the answer - NSC Physical Sciences - Question 8 - 2017 - Paper 1

Bulb R is brighter than bulb P because in this arrangement, bulbs P and R are parallel to each other while being part of a single circuit branch. Since R has the same potential difference across it as P but carries twice the current (due to the combined paths), it emits more light, making it brighter.

Bulb T does not light up at all because the wire connecting it effectively shorts it out. This means that the majority of current bypasses bulb T, leading to zero current and thus no brightness. In contrast, bulb R will shine brightly as it receives the full current.

To find the current in the 8 Ω resistor, we first calculate the total resistance in the circuit. The equivalent resistance (R_eq) of the 5 Ω and 10 Ω resistors in series is:

$R_{eq} = R_1 + R_2 = 5 \, \Omega + 10 \, \Omega = 15 \, \Omega.$

Adding the internal resistance of the battery gives us:

$R_{total} = R_{eq} + R_{internal} = 15 \, \Omega + 1 \, \Omega = 16 \, \Omega.$

Using Ohm's law, the total current (I) supplied by the battery (20V) can be calculated as:

$I = \frac{V}{R_{total}} = \frac{20 \, V}{16 \, \Omega} = 1.25 \, A.$

Therefore, the current in the 8 Ω resistor, assuming it carries the full current, is also 1.25 A.

The voltage across the 5 Ω resistor can be calculated using Ohm’s law:

$V = I \times R = 1.25 \, A \times 5 \, \Omega = 6.25 \, V.$

Thus, the potential difference across the 5 Ω resistor is 6.25 V.

The total power (P) supplied by the battery can be calculated using the formula:

$P = I^2 \times R_{total} = (1.25 \, A)^2 \times 16 \, \Omega = 20 \, W.$

Thus, the total power supplied by the battery is 20 W.