Series Circuit with a Lamp and Internal Resistance
A student in a Herat physics lab connects a battery of emf 12 V and internal resistance 1.5 ohm in series with a lamp rated 6 V, 3 W and a fixed resistor R. The lamp is designed to operate at its rated values (6 V across it, 3 W dissipated). The student wants the lamp to glow at exactly its rated power. Assume the lamp behaves as an ohmic resistor at its rated operating point. The circuit is a single series loop: battery, lamp, and resistor R. The student measures the current with an ammeter and finds it is not at the design value, so the resistor value must be chosen correctly.
1. Determine the rated current and the resistance of the lamp, then find the value of R that makes the lamp operate at its rated power.
Read the worked answer
Rated current: P = VI, so I = P/V = 3/6 = 0.5 A. Lamp resistance: R_lamp = V/I = 6/0.5 = 12 ohm (or P = I^2 R gives R = 3/0.25 = 12 ohm).
In series, the same current 0.5 A flows everywhere. Total emf must equal the sum of voltage drops plus the internal drop: 12 = V_lamp + V_R + Ir 12 = 6 + 0.5R + 0.5*1.5 12 = 6 + 0.5R + 0.75 0.5R = 12 - 6.75 = 5.25 R = 10.5 ohm.
2. The student accidentally uses R = 6 ohm. Calculate the new current and explain whether the lamp will be brighter or dimmer than at its rated point.
Read the worked answer
Total resistance = R_lamp + R + r = 12 + 6 + 1.5 = 19.5 ohm. Current I = emf / R_total = 12 / 19.5 = 0.615 A (approximately).
This current (0.615 A) is greater than the rated 0.5 A. Power in the lamp = I^2 * R_lamp = (0.615)^2 * 12 = 0.378 * 12 = 4.54 W, which exceeds the rated 3 W. So the lamp glows brighter than rated and may be damaged. (Note: the assumption of constant lamp resistance is only an approximation; in reality the filament heats up and resistance rises, but the qualitative conclusion - brighter, over-rated - holds.)
3. Explain why the internal resistance reduces the terminal voltage of the battery as current increases, and compute the terminal voltage when R = 10.5 ohm.
Read the worked answer
The terminal voltage is V_terminal = emf - Ir. As current I increases, the voltage dropped across the internal resistance (Ir) increases, so the terminal voltage available to the external circuit decreases.
With R = 10.5 ohm, I = 0.5 A. Then V_terminal = 12 - 0.51.5 = 12 - 0.75 = 11.25 V. This equals the sum of the external drops: V_lamp + V_R = 6 + 0.510.5 = 6 + 5.25 = 11.25 V, confirming consistency.