Two cells of EMF $1.5\,\text{V}$ and $2\,\text{V}$ and internal resistances $1\,\Omega$ and $2\,\Omega$ respectively are connected in parallel to pass current in the same direction through an external $5\,\Omega$ resistor. (i) Draw the circuit diagram. (ii) Using Kirchhoff's laws, find the current through each branch and the potential difference across the $5\,\Omega$ resistor.

Current through the $5\,\Omega$ resistor $=\dfrac{5}{17}\,\text{A}\approx0.29\,\text{A}$; potential difference across it $=\dfrac{5}{17}\times5\approx1.47\,\text{V}$. (Branch currents: $I_1\approx0.03\,\text{A}$ through the $1\,\Omega$ cell, $I_2\approx0.26\,\text{A}$ through the $2\,\Omega$ cell.)

CBSE Class 12 · 2005 · All India · Current ElectricityShort Answer

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Two cells of EMF

1.5 V

and

2 V

and internal resistances

1 Ω

and

2 Ω

respectively are connected in parallel to pass current in the same direction through an external

5 Ω

resistor.
(i) Draw the circuit diagram.
(ii) Using Kirchhoff's laws, find the current through each branch and the potential difference across the

5 Ω

resistor.

Solution

Let

I_{1}, I_{2}

be the currents from the two cells and

I_{1} + I_{2}

the current through the

5 Ω

resistor. Applying the loop rule to the two loops gives

2 I_{2} - I_{1} = 0.5

and

6 I_{1} + 5 I_{2} = 1.5

. Solving,

I_{2} = \frac{9}{34} A

and

I_{1} = \frac{5}{170} A

, so the current through the

5 Ω

resistor is

I_{1} + I_{2} = \frac{5}{17} A

and the PD across it is

(\frac{5}{17}) \times 5 = 1.47 V .

CBSE Class 12 Board · 2005 · All India · Subjective

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CBSE Class 12 · 2005 · All India · Current ElectricityShort Answer

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