Question

2 Batteries who EMF is 3 volts and internal resistance is 2 Ω are connected in parallel combination. These two batteries are connected with a resistance of 2 Ω. Find the current in the circuit.​

Answer 1

Question :-

→ Given above .

Answer :-

→ Current flow in both internal resistances is 0.5 ampere ,but current flow in external resistance (2 ohm ) is 1 amp . and potential difference across external resistance is 2 volt .

Explanation :-

$\bf \: let \: i_1 \: , i_2 \: and \: i_1 \: + \: i_2 \: \: is \: current \: flow \: in \\ \bf \: (internal \: resistance) \: 2 \Omega\: , 2\Omega \: and \: 2 \Omega \\ \bf (external \: resistance \: ) \\ \\ \bf apply \: kvl \: in \: closed \: circuit \: CFEDC \: \\ \\ \to \bf \: - 2i_1 + 2i_2 - 3 + 3 = 0 \\ \\ \to \bf \: 2i_2 - 2i_1 = 0 \\ \\ \to \boxed{ \bf i_2 = i_1} \: \: ...eq.(1) \\ \\ \bf now \: again \: apply \: kvl \: in \: CDABC \: \\ \\ \to \bf \: - 2i_2 - 2(i_1 + i_2) + 3 = 0 \\ \\ \to \bf \: - 2i_1 -4 i_2 + 3 = 0 \\ \\ \bf from \: \: eq.(1) \\ \\ \to \bf - 2i_2 - 4i_2 + 3 = 0 \\ \\ \to \bf \: - 6i_2 = - 3 \\ \\ \mapsto \boxed{\bf \: i_2 =i_1 = \frac{1}{2} \: = 0.5 \: amp.}$

Now ,

$\bf \: current \: through \: 2 \Omega\: (external \: resistance \: ) \\ \\ \to \bf i_1 + i_2 \\ \\ \to \bf \: \frac{1}{2} + \frac{1}{2} = 1 \: amp. \\ \bf hence \: potential \: across \: 2 \Omega \: \\ \bf \: resistance \: is \: \\ \\ \mapsto \bf v = 1 \times 2 \\ \\ \mapsto \bf \: v = 2 \: volt$

Answer 2

$\mathrm{ \fcolorbox{green}{lightgreen} {Verified \: answer}}$

The current is 2 volt.

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