ICSE Class 10 Answered
fig. 8.37 below shows a battery of emf 9V & internal resistance 0.6ohm, connected to three resisters A, B & C. Calculate:
a) the combined resistance of B & C.
b) the total resistance of A, B & C.
c) the total resistance of the circuit,
d) the current in each resistor A, B & C. (please explain this question with explaination as i have doubt in this one)
e) the potential drop across the internal resistance,
f) the potential difference across the resistor B,
g) the terminal voltage of the cell.
Asked by nethra | 04 Mar, 2018, 12:28: PM
Expert Answer
(a) combined resistance of B and C which are parallel : (6×4)/(6+4) = 2.4Ω
(b) total reswistance A, B and C = 2+2.4 = 4.4
(
(c) total resistance of the circuit = 2+2.4+0.6 = 5 (internal resistance of battery 0.6 is added)
(d) current in each resistor:-
current in the circuit = 9/5 =1.8 A
same 1.8A will flow through internal resistance, Resitor A and parallel combination of B and C.
Current through Resistance B = 1.8 × (6/10) = 1.08 A
Current through Resistance C = 1.8 × (4/10) = 0.72 A
(e) potential drop across inernal resistance = 0.6×1.8 = 1.08 V
(f) potential drop across resistor B = 1.08×4 = 4.32
(g) terminal voltage = 9 - 1.08 = 7.92 V
Answered by Thiyagarajan K | 04 Mar, 2018, 06:18: PM
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