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Two Resistances R1 = 4ω and R2 = 6ω Are Connected in Series. the Combination is Connected with a Battery of E.M.F. 6v and Negligible Resistance. Calculate: - Physics

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प्रश्न

Two resistances R1 = 4Ω and R2 = 6Ω are connected in series. The combination is connected with a battery of e.m.f. 6V and negligible resistance. Calculate:
(i) the heat produced per minute in each resistor,
(ii) the power supplied by the battery.

टिप्पणी लिखिए

उत्तर

Given: R1 = 4Ω, R2 = 6Ω, E = 6V, t = 1 minute = 60s.
Total resistance of the circuit R = R+ R2 = 4 + 6 = 10Ω

Current through the battery I = `"E"/"R" = 6/10 = 0.6` A

The same current flows through each resistor.

(i) The heat produced in resistor R1 is H1 = I2R1t

= (0.6)2 × 4 × 60 = 86.4 J

The heat produced in resistor R2 is H2 = I2R2

= (0.6)2 × 6 × 60 = 129.6 J

(ii) The power supplied by the battery = E × I 

= 6 × 0.6 = 3.6 W.

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अध्याय 8: Electric Power and House hold Circuits - Long Numericals

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आईसीएसई Physics [English] Class 10
अध्याय 8 Electric Power and House hold Circuits
Long Numericals | Q 2

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