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Asked by vidyavikram10 | 02 Mar, 2020, 17:17: PM
Expert Answer
Figure shows the given circuit that includes the internal resistance of batteries .
Current in the circuit is calculated using Kirchoff's voltage law that is applied to the given loop.
Let i be the current flowing in the circuit. Then we have,
( 0.5 i + 0.5 i + 4 i + 6 + i ) V = 12 V or i = 1 A
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(1) potential at point A is voltage drop acorss 4Ω resistor, i.e., 4 Ω × 1 A = 4 V
(2) potential at point B is voltage drop acorss 0.5 Ω resistor between B and C, i.e., 0.5 Ω × 1 A = 0.5 V
(3) ideal voltmeter connected across 6V battery will read the EMF of battery and voltage drop across internal resistance .
This is because 6V battery is charged by current in the circuit . Hence voltmeter reading = 6 V + ( 1Ω × 1 A ) = 7V
(4) as explained in part (3)
Answered by Thiyagarajan K | 02 Mar, 2020, 20:28: PM
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