Answer to Question #125654 in Electric Circuits for Vidurjah Perananthan

Question

Suppose you have to connect a battery and four equal light bulbs, A, B, C and D.
Lamp A and B should shine just as brightly, C should shine brighter, and D should shine most brightly. Draw a wiring diagram and motivate your choice

Answer 1

Explanations & Calculations

• Refer to the sketch
• Both the circuits serve the purpose. But in the second circuit, the total circuit resistance is lower than that in the first one.
• Then, the second circuit draws much amount of current compared to the first one.
• Then each of the bulb receives a greater fraction of the total amount compared to the individual currents each receives in the first arrangement.
• Therefore, each of the bulb in the second arrangement shines brighter than in the first one while serving the given requirement.

Proof

• If the resistance of a bulb is R then,

1. The total resistance of the first arrangement = $\small \frac{5R}{3}$

• Current drawn from the battery = $i_1$ = $\small \frac{3}{5}\big(\frac{E}{R}\big)$
• Current through the $\small 1^{st}$ bulb = $\small \frac{3}{5}\big(\frac{E}{R}\big)$
• Current through the $\small 2^{nd}$ bulb = $\small \frac{2i_1}{3}$ = $\small \frac{2}{5}\big(\frac{E}{R}\big)$
• Current through 3/4 bulbs = $\small\frac{i_1}{3}$ = $\small \frac{1}{5}\big(\frac{E}{R}\big)$

2, Total resistance of the second arrangement = $\small \frac{3R}{5}$

• Current drawn from the battery = $i_2$ = $\small \frac{5}{3}\big(\frac{E}{R}\big)$
• Current through the $\small 1^{st}$ bulb = $\small \big(\frac{E}{R}\big)$
• Current through the $\small 2^{nd}$ bulb = $\small \frac{2}{3}\big(\frac{E}{R}\big)$
• Current through 3/4 bulbs = $\small \frac{1}{3}\big(\frac{E}{R}\big)$