हिंदी

A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?

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प्रश्न

A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?

संख्यात्मक
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उत्तर

When equivalent resistances are connected in series,

R = R1 + R2 + R3 + R4 + R5

R1 = 0.2 Ω, R2 = 0.3 Ω, R3 = 0.4 Ω, R= 0.5 Ω, R5 = 12 Ω  ....(Given)

R = 0.2 + 0.3 + 0.4 + 0.5 + 12

R = 13.4 Ω

V = IR  ....(According to Ohm’s law)

I = `"V"/"R"`

= `9/13.4`

= 0.67 A

There will be no division of electric current because the resistors are connected in series.

0.67 A of electric current will flow in a 12 Ω resistor.

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अध्याय 11: Electricity - EXERCISE [पृष्ठ १९३]

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एनसीईआरटी Science [English] Class 10
अध्याय 11 Electricity
EXERCISE | Q 9. | पृष्ठ १९३

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