The dry cell of E-M-F 12 V- is connected in series to a 3 - and a 6 - resistor- Calculate the heat energy produced by the 6- resistor in 5 minutes-

The correct option is B 3200 J Total Resistance R_T=3+6=9 Current, i=VR_T=129=43A Power across 6Ω: P=i^2R=169× 6=323W Heat energy prod ...

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Byju's Answer

Standard X

Physics

Electric Power

The dry cell ...

Question

The dry cell of E.M.F $12 V$ , is connected in series to a $3 Ω$ and a $6 Ω$ resistor. Calculate the heat energy produced by the $6 Ω$ resistor in 5 minutes.

50 J

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3200 J

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(480 \times 360) J

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3000 J

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Solution

The correct option is B $3200 J$
Total Resistance $R_{T} = 3 + 6 = 9$

Current, $i = \frac{V}{R_{T}} = \frac{12}{9} = \frac{4}{3} A$
Power across $6 Ω :$ $P = i^{2} R = \frac{16}{9} \times 6 = \frac{32}{3} W$

Heat energy produced, $E = P \times t = \frac{32}{3} \times 5 \times 60 = \frac{9600}{3} J = 3200 J$

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The dry cell of E.M.F 12 V, is connected in series to a 3Ω and a 6Ω resistor. Calculate the heat energy produced by the 6Ω resistor in 5 minutes.

The correct option is B 3200JTotal Resistance RT=3+6=9Current, i=VRT=129=43APower across 6Ω:P=i2R=169×6=323WHeat energy produced, E=P×t=323×5×60=96003J=3200J

The dry cell of E.M.F $12 V$ , is connected in series to a $3 Ω$ and a $6 Ω$ resistor. Calculate the heat energy produced by the $6 Ω$ resistor in 5 minutes.