When switch S in the figure is open, the voltmeter V of the battery reads 3.11 V . When the switch is closed, the voltmeter reading drops to 3.00 V , and the ammeter A reads 1.63 A . Assume that the two meters are ideal, so they do not affect the circuit. (Figure 1) Find the internal resistance r of the battery.
When switch S in the figure is open, the voltmeter V of the battery reads 3.11 V . When the switch is closed, the voltmeter reading drops to 3.00 V , and the ammeter A reads 1.63 A . Assume that the two meters are ideal, so they do not affect the circuit. (Figure 1)
Find the internal resistance r of the battery.
Answer:-
Given
The Resistor R and the internal resistance r are connected in series when the switch is closed.
The current passing through the circuit when switch is open then
I = 0
The voltage of the battery
V = E - I r ( E = emf)
V = E - 0 × r
E = V
E = 3.11 V
From the circuit the terminal voltage is
Vt = E - I r
I r = E - Vt
r = (E - Vt)/I
Internal resistance(r) = (3.11 - 3.00)/(1.63)
r = (0.11)/(1.63)
r = 0.0675 ohms
Trending now
This is a popular solution!
Step by step
Solved in 2 steps
