The dry cell of E-M-F 12 V- is connected in series to a 3 Omega and a 6 Omega resistor- Calculate the heat energy produced by the 6Omega resistor in 5 minutes-50 J3200 J3000 J-480times 360- J

Question

Toppr Admin · Accepted Answer

Total Resistance RT-3-6-9Current- i-VRT-129-43APower across 6-x3A9-P-i2R-169-xD7-6-323WHeat energy produced- E-P-xD7-t-323-xD7-5-xD7-60-96003J-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.

$50 J$
$3200 J$
$3000 J$
$(480 \times 360) J$

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

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.50J3200J3000J(480×360)J

Total 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.

$50 J$
$3200 J$
$3000 J$
$(480 \times 360) J$

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