Ncert Exercises & Solutions

Two cells of voltage 10 V and 2 V and internal resistances 10 $Ω$ $Ω$ and 5 $Ω$ $Ω$ respectively, are connected in parallel with the positive end of the 10 V battery connected to the negative pole of the 2V battery (figure). Find the effective voltage and effective resistance of the combination.

Hint: Apply Kirchoff's laws.

Step 1: Applying Kirchhoff's junction rule,

I_{1} = I + I_{2}

$I_{1} = I + I_{2}$

Applying Kirchhoff's IInd law/loop rule applied in outer loop containing 10 V cell and resistance R, we have;

10 = IR + 10 I_{1} . . . (i)

$10 = IR + 10 I_{1} . . . (i)$

Step 2: Applying Kirchhoff IInd law/loop rule applied in upper loop containing 2 V cell and resistance R, we have;

2 = 5 I_{2} - RI = 5 (I_{1} - I) - RI

$2 = 5 I_{2} - RI = 5 (I_{1} - I) - RI$

or 4 = 10 I_{1} - 10 I - 2 RI . . . (ii)

$or 4 = 10 I_{1} - 10 I - 2 RI . . . (ii)$

Solving Eqs. (i) and (ii),

\Rightarrow 6 = 3 RI + 10 I

$\Rightarrow 6 = 3 RI + 10 I$

2 = I (R + \frac{10}{3})

$2 = I (R + \frac{10}{3})$

Step 3: Also, the external resistance is R. Ohm's law states that;

{V=I(R+R}_{eff})

${V=I(R+R}_{eff})$

On comparing, we have V= 2 V and effective internal resistance;

(R_{elf}) = (\frac{10}{3}) Ω

$(R_{elf}) = (\frac{10}{3}) Ω$

Since the effective internal resistance (

R_{eff}

$R_{eff}$ ) of two cells is

(\frac{10}{3}) Ω

$(\frac{10}{3}) Ω$ being the parallel combination of 5

Ω

$Ω$ and 10

Ω

$Ω$ . The equivalent circuit is given below;

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