91影视

Power is defined as the energy per unit of time and can be expressed as,

$\mathit{P}\mathbf{=}{\mathit{I}}^{\mathbf{2}}\mathit{R}\mathbf{}\mathbf{=}\frac{{\mathbf{V}}^{\mathbf{2}}}{\mathbf{R}}$鈥︹赌︹赌︹赌︹赌︹赌︹赌︹赌�(1)

two identical bulbs are connected in parallel.

E = 8.0 V

internal resistance 0.8 鈩�

resistance R = 2.0 鈩�

(a)

Current flowing each bulb in the circuit,

$I=\frac{E}{R_{eq.}+R_{int}}$ (2)

Also,

$$\begin{gathered} \frac{1}{R_{eq.}}=\frac{1}{R}+\frac{1}{R} \\ {R}_{eq.}=\frac{R}{2} \\ {R}_{eq.}=\frac{\text{2.0}惟}{2} \\ {R}_{eq.}=1.0惟 \end{gathered}$$

Put these values in equation (2)

$$\begin{gathered} I=\frac{E}{R_{eq.}+R_{int}} \\ =\frac{8.0V}{1.0惟+0.8惟} \\ =4.4A \end{gathered}$$

The voltage across the two bulbs is the same because they are connected in parallel. Resistance is also the same, current flowing is the same. The total current is given as,

$$\begin{gathered} I=I_R+I_R \\ I=2I_R \\ {I}_R=\frac{4.4A}{2} \\ {I}_R=2.2A \end{gathered}$$

Potential is,

$$\begin{gathered} V_R=I_{R}R \\ =(2.2A)(2.0惟) \\ =4.4V \end{gathered}$$

Power delivered to each bulb,

$$\begin{gathered} P_R=I_R^{2}R \\ =(2.2A{)}^2(2.0惟) \\ =9.7W \end{gathered}$$

Thus, the current through each bulb, the potential difference across each bulb, and the power delivered to each bulb are 2.2 A, 4.4 V, and 9.7 W respectively.

(b)

Take that one bulb burns out and filament breaks due to which currently do not flow through it which can be given as,

$$\begin{gathered} I=\frac{E}{R_{eq.}+R_{int}} \\ =\frac{8.0V}{2.0惟+0.8惟} \\ =2.9A \end{gathered}$$

Power delivered,

$$\begin{gathered} P=I^{2}R \\ =(2.9A{)}^2(2.0惟) \\ =17W \end{gathered}$$

Thus, the power delivered to the remaining bulb is P = 17 W. the remaining bulb glows more or less brightly after the other bulb burns out than before because it consumes more power.