Home›MPPSC Forest›State Forest Service Mains Exam›Section 'B': Forestry and General Science›UNIT-07: Physics›Electricity and its effects: Electric intensity, potential, potential difference, electric current, Ohm's law, resistance, specific resistance, factors influencing resistance, combination of resistance (numerical examples), thermal effect of current and its use, calculation of power and electrical energy spent (numerical), precautions in electric experiments›A battery of e.m.f. 1.5 V has a terminal potential difference of 1.25

A battery of e.m.f. 1.5 V has a terminal potential difference of 1.25 V when an external resistance of 25 Ω is connected to it. The internal resistance of the battery will be

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A.25 Ω
B.15 Ω
C.5 ΩCorrect
D.0 Ω
Explanation & Solution

Correct Answer: (C) 5 Ω

The internal resistance r of a cell is given by r = R * ((E / V) - 1). Substituting E = 1.5 V, V = 1.25 V, and R = 25 Ω yields r = 25 * ((1.5 / 1.25) - 1) = 25 * (1.2 - 1) = 25 * 0.2 = 5 Ω.

हिंदी में प्रश्न एवं आदर्श उत्तर

1.5 V ई०एम०एफ० की बैटरी में 25 Ω का बाहरी प्रतिरोध जुड़नेपर टर्मिनल विभवांतर 1.25 V होता है। बैटरी का आंतरिक प्रतिरोध होगा

A.25 Ω
B.15 Ω
C.5 Ωसही
D.0 Ω

सही उत्तर: (C) 5 Ω

सेल का आंतरिक प्रतिरोध r सूत्र r = R * ((E / V) - 1) द्वारा दिया जाता है। E = 1.5 V, V = 1.25 V और R = 25 Ω रखने पर r = 25 * ((1.5 / 1.25) - 1) = 25 * (1.2 - 1) = 25 * 0.2 = 5 Ω प्राप्त होता है।

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