Find the emf (E) of the cell in V required to | vedclass

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(D) $1$. The bulb is rated $P = 4.5 \, W$ and $V = 1.5 \, V$. The current through the bulb is $I_b = P / V = 4.5 / 1.5 = 3 \, A$.$2$. The bulb is in p

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$13.5$

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(D) $1$. The bulb is rated $P = 4.5 \, W$ and $V = 1.5 \, V$. The current through the bulb is $I_b = P / V = 4.5 / 1.5 = 3 \, A$.$2$. The bulb is in parallel with a resistor of $R = 1 \, \Omega$. The voltage across the resistor is the same as the bulb,$V = 1.5 \, V$. The current through the resistor is $I_r = V / R = 1.5 / 1 = 1.5 \, A$.$3$. The total current flowing through the cell is $I = I_b + I_r = 3 \, A + 1.5 \, A = 4.5 \, A$.$4$. The cell has an internal resistance $r = 2.67 \, \Omega$. The $emf$ of the cell is given by $E = V + I \cdot r$.$5$. Substituting the values: $E = 1.5 + (4.5 	imes 2.67) = 1.5 + 12.015 = 13.515 \, V$. Rounding to the nearest option,$E = 13.5 \, V$.

Find the $emf$ $(E)$ of the cell in $V$ required to light the bulb rated $4.5 \, W, 1.5 \, V$ fully,as shown in the circuit.

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