Figure (32-E2) shows an arrangement to measure the emf ε and internal resistance r of a battery. The voltmeter has a very high resistance and the ammeter also has some resistance. The voltmeter reads 1.52 V when the switch S is open. When the switch is closed, the voltmeter reading drops to 1.45 V and the ammeter reads 1.0 A. Find the emf and the internal resistance of the battery.
Figure 32-E2
Answers
Answer:
When the switch S is opened, loop 2 will be open and no current will pass through the ammeter. Loop 1 will be closed; so, a current will flow through it. But since the voltmeter has very high resistance, compared to the internal resistance of the battery, the voltage-drop across the internal resistance can be ignored, compared to the voltage drop across the voltmeter. So, the voltage appearing across the voltmeter will be almost equal to the emf of the battery.
∴ ε = 1.52 V
(b) When the switch is closed, current will pass through the circuit in loop 2. In this case, there will be a voltage drop across r due to current i flowing through it.
Applying the loop rule, we get:
ε – ir = 1.45
⇒ 1.52 – ir = 1.45
⇒ ir = 0.07
⇒ 1.r = 0.07
r = 0.07 Ω
The emf and the internal resistance of the battery is 0.07 Ω.
Explanation:
From the given figure when switch is open no current will pass through the ammeter. Hence loop 1 will be closed and current will flow through it.
But voltmeter has very high resistance than the resistance of the battery. Due to high value of resistance the voltage appearing across the voltmeter will be equal to the emf of the battery.
∴ ε = 1.52 V
Step 1:
Considering the other way, when switch closes, current will pass through the circuit in loop 2. There will be a voltage drop across resistor (r) due to current (i) flowing through it.
Step 2:
After applying the loop rule,
ε – ir = 1.45
⇒ 1.45 = 1.52 – ir
⇒ ir = 0.07
⇒ ir = 0.07
Finally, Resistance of the battery (R) = 0.07 Ω
