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Question
Two cells of voltage 10V and 2V and internal resistances 10Ω and 5Ω respectively, are connected in parallel with the positive end of 10V battery connected to negative pole of 2V battery (Figure). Find the effective voltage and effective resistance of the combination.
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Solution
In this problem first, we are applying Kirchhoff's junction rule at c, I1 = I + I2
Applying Kirchhoff's voltage law in loop (e – f – b – a – e) loop L1 outer loop, then we get
10 – IR – 10l1 = 0
10 = IR + 10l1 ......(i)
Applying Kirchhoff's voltage law in loop (c – b – a – d – c) loop L2, we get
– 2 – IR + 5I2 = 0
2 = 5I2 – RI
As we know, I1 = I + I2 then
I2 = I1 – I
So the above equation can be written as
2 = 5(I1 – I) – RI
or 4 = 10I1 – 10I – 2RI ......(ii)
Subtracting equations (ii) from (i), we get
⇒ 6 = 3RI + 10I
2 = `I(R + 10/3)`
Also, the external resistance is R. The Ohm's law states that V = I(R + Reff)
On comparing, we have V = 2V and effective internal resistance `(R_("eff")) = (10/3)Ω`
Since, the equivalent internal resistance (Reff) of two cells is `(10/3)Ω`, being the parallel combination of 5Ω and 10Ω. The equivalent circuit is given below:
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