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1 answer:
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Answer:
The current is reduced to half of its original value.
Explanation:
- Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:
- where Rint = r and RL = r
- Replacing these values in I₁, we have:
- When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:
- We can find the relationship between I₂, and I₁, dividing both sides, as follows:
- The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.
Answer:
Uncorrected values for
For circuit P
R = 2.4 ohm
For circuit Q
R = 2.4 ohm
Corrected values
for circuit P
R = 12 OHM
For circuit Q
R = 2.3 ohm
Explanation:
Given data:
Ammeter resistance 0.10 ohms
Resister resistance 3.0 ohms
Voltmeter read 6 volts
ammeter reads 2.5 amp
UNCORRECTED VALUES FOR
1) circuit P
we know that IR =V
2) circuit Q
R = 2.4 ohm as no potential drop across ammeter
CORRECTED VALUES FOR
1) circuit p
IR = V
R= 12 ohm
2) circuit Q
R = 2.3 ohm
