The Switch That Changed the Circuit
A student builds a circuit using a 6.0 V battery of negligible internal resistance, a 10 Ω resistor R1, a 15 Ω resistor R2, an ammeter that reads the current leaving the battery, and a voltmeter connected across R2. With switch S open, the current leaves the battery, passes through R1 and then through R2 in a single loop back to the battery. Closing S changes only the connections: R1 and R2 then each form their own separate path directly between the two battery terminals. The resistors are ohmic and their temperature does not change appreciably.
1. Calculate the ammeter and voltmeter readings with S open.
Read the worked answer
With S open the resistors are in series, so the total resistance is 10 + 15 = 25 Ω.
Ammeter: I = V/R = 6.0/25 = 0.24 A.
Voltmeter across R2: V₂ = I × R2 = 0.24 × 15 = 3.6 V.
As a check, the share taken by R1 is 0.24 × 10 = 2.4 V, and 2.4 + 3.6 = 6.0 V, the battery voltage. In a series circuit the larger resistance takes the larger share of the voltage.
2. Calculate both readings with S closed, and state whether the voltmeter reading rises or falls.
Read the worked answer
With S closed the resistors are in parallel across the battery.
Combined resistance: 1/R = 1/10 + 1/15 = 3/30 + 2/30 = 5/30, so R = 6.0 Ω.
Ammeter: I = 6.0/6.0 = 1.0 A.
Voltmeter across R2: R2 is now connected directly to the battery terminals, so V₂ = 6.0 V.
The reading rises, from 3.6 V to 6.0 V. The branch currents are 0.6 A through R1 and 0.4 A through R2, which add to the 1.0 A main current.
3. The resistance of R1 has not changed, yet the current through it is different after S is closed. Explain why, and state whether the total power drawn from the battery increases or decreases.
Read the worked answer
With S open, R1 shares the 6.0 V with R2 and receives only 2.4 V, so its current is 0.24 A. Closing S puts R1 directly across the battery, so the full 6.0 V now drives it and its current becomes 6.0/10 = 0.60 A. The resistor is unchanged; the voltage across it is what changed, and current is proportional to that voltage.
Total power: S open, P = 6.0 × 0.24 = 1.44 W; S closed, P = 6.0 × 1.0 = 6.0 W. The power increases, because the total resistance fell from 25 Ω to 6 Ω and the battery supplies a much larger current.