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Question

# A battery of emf 12V and internal resistance 2Ω is connected with two resistors A and B of resistance 4Ω and 6Ω respectively joined in series. Find:(i) Current in the circuit. (ii) The terminal voltage of the cell. (iii) The potential difference across 6Ω Resistor. (iv) Electrical energy spent per minute in 4Ω Resistor.

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Solution

## Step 1: Make circuit diagram and apply KVL [Refer Fig. 1]While applying KVL in anti-clockwise direction, increase in voltage is considered as positive and decrease in voltage as negative. 12−4i−6i−2i=0 ⇒ I=1A Step 2: Find terminal voltage of the cell [Refer Fig. 2] Vterminal=E−ir =12−2i=12−2×1=10 VStep 3: Apply Ohm's Law for 6Ω Resistance [Refer Fig. 3] VB=iRB=6i=6×1=6 VStep 4: Energy spent in 4Ω Resistance [Refer Fig. 4]Energy Spent in 1 minute(t=60 sec): EA=i2RAt=(1A)2×4Ω×60s =240J

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