Question 12
Equivalent resistance of the circuit
R = 4 + [(3×6)/(3+6)] Ohm
= 6 Ohm
Question 13
Current in the circuit I = V/R
= 12/6 A
= 2 A
Potential drop across 4 Ohm resistance is V' = 2×4 Volt
= 8 Volt
Now potential difference between X and Y is
VXY = 12 - 8 volt
= 4 Volt
Question 14
Since the Second bulb is connected in parallel, so there will be no change in current in the circuit, hence ammeter reading. Also voltage will also be the same, so voltmeter reading will not change.
# Last option is correct.
12-16 an 3 of 8 12V in un The total equivalent resistance of the circuit shown...
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Consider the circuit shown below. All three batteries are ideal. The voltmeter and ammeter are also ideal. 7.00V 崇弓= 12.0V v) R 5.00 R2 6.00 Ω = 8.00 V A. What is the internal resistance of the batteries? B. What is the internal resistance of the voltmeter? C. what is the internal resistance of the ammeter? D. Based on your answers to questions A-C draw a simpler equivalent circuit (in the space above to the right of the original circuit)...
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