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Mixed Combination of Cells

In the circui...

Question

In the circuit in figure, we have $R_{2} = 20 Ω$ , $R_{3} = 15 Ω$ , and the current flowing through resistor $R_{2}$ is 0.3 A. The ammeter shows 0.8 A. Find the resistance $R_{1}$ :

20 Ω

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30 Ω

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40 Ω

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60 Ω

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Solution

The correct option is D $60 Ω$
Potential drop across $R_{2}$ = Potential drop across $R_{1}$ = Potential drop across $R_{3}$
Considering $R_{2}$ and $R_{3}$
$i_{2} R_{2} = i_{3} R_{3}$
$0.3 \times 20 = 15 \times i_{3}$
$i_{3} = 0.4 A$
Current passing through ammeter $= i_{1} + i_{2} + i_{3} = 0.8$
$i_{1} + 0.3 + 0.4 = 0.8$
$i_{1} = 0.1 A$
PD across $R_{1}$ should be equal to the potential drop across $R_{2}$
$i_{1} R_{1} = i_{2} R_{2}$
$0.1 R_{1} = 0.3 \times 20$
$R_{1} = 60 o h m$
Hence option D is correct.

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Similar questions

Resistors $R_{1}$ =10 $Ω$ , $R_{2}$ = 40 $Ω$ , $R_{3}$ = 30 $Ω$ , $R_{4}$ = 20 $Ω$ , $R_{5}$ = 60 $Ω$ and a 12V battery is connected as shown.

Calculate:
(a) the total resistance
(b) the total current flowing in the circuit.

R_{1} = 10 Ω

R_{2} = 40 Ω

R_{3} = 30 Ω

R_{4} = 20 Ω

R_{5} = 60 Ω

R_{1}

R_{1} = 10 Ω, R_{2} = 20 Ω, R_{3} = 30 Ω

V_{1} = 10 V, V_{2} = 6 V

V_{3} = 5 V

R_{3}

(R_{1} = 15 Ω, R_{2} = 15 Ω, R_{3} = 30 Ω, R_{4} = 35 Ω)

20 Ω

0.3 A

0.8 A

R_{1}

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