The dry cell of E-M-F 12 V- is connected in series to a 3 Omega and a 6 Omega resistor- Calculate the total power supplied by the dry cell-72 W16 W9 W2 W

Question

Toppr Admin · Accepted Answer

Equivalent resistance for resistance connected in series is given by

$R_{e q} = R_{1} + R_{2}$

Since, in the given case $3 Ω$ and $6 Ω$ are connected in series.
Hence, equivalent resistance, $R_{e q} = 3 + 6 = 9 Ω$

Given, emf of cell $V = 12 V$

Hence, the power supplied by the cell is given by, $P = \frac{V^{2}}{R}$

$P = \frac{12 \times 12}{9} = 16 W$

The dry cell of E.M.F 12 V, is connected in series to a 3Ω and a 6Ω resistor. Calculate the total power supplied by the dry cell.72 W16 W9 W2 W

Equivalent resistance for resistance connected in series is given by Req=R1+R2Since, in the given case 3 Ω and 6 Ω are connected in series.Hence, equivalent resistance, Req=3+6=9 ΩGiven, emf of cell V=12 VHence, the power supplied by the cell is given by, P=V2RP=12×129=16 W

The dry cell of E.M.F $12 V$ , is connected in series to a $3 Ω$ and a $6 Ω$ resistor. Calculate the total power supplied by the dry cell.

$72 W$
$16 W$
$9 W$
$2 W$