91Ó°ÊÓ

According to Kirchhoff's loop rule, the total of all electric potential differences within a loop is zero. Kirchhoff's voltage law or Kirchhoff's second law are other names for it. Given that energy cannot enter or leave a closed circuit, this means that the battery's energy is consumed by all of the other components in a loop.

• Voltage of cells:\(1.58{\rm{ }}V\)
• Internal resistance of cell:\(0.0200{\rm{ }}\Omega \)
• Voltage of carbon-zinc cell:\(1.53{\rm{ }}V\)
• Internal resistance of carbon-zinc cell:\(0.100{\rm{ }}\Omega \)
• The load resistance is: \(10{\rm{ }}\Omega \)

a)

Circuit diagram of the toy and its batteries is

b)

Voltage is equal to:

\(E = 3 \times 1.58 + 1.53 = 6.27\;V\)

Internal resistances are equal to:

\(R = 3 \times 0.02 + 0.1 = 0.16{\rm{ }}\Omega \)

Calculating current flow

\(\begin{aligned}{c}I &= \frac{E}{{{R_t}}}\\I = \frac{{6.27}}{{0.16 + 10}}\\I &= 4.89\;A\end{aligned}\)

Therefore, the current value is \(I = 4.89\;A\).

c)

Power supplied is equal to:

\(\begin{aligned}{c}P = R{I^2}\\P = 10 \times {(4.89)^2}\\P = 239\;W\end{aligned}\)

Therefore, value of power value is \(P = 239\;W\).

d)

In case the battery goes bad, its resistance can be calculated as:

\(\begin{aligned}{c}{I_2} &= \sqrt {\frac{{{P_2}}}{R}} &= \sqrt {\frac{{0.5}}{{10}}} &= 0.223\;A\\{R_{t2}} &= \frac{E}{{{I_2}}} &= \frac{{6.27}}{{0.223}} &= 28.12{\rm{ }}\Omega \end{aligned}\)

Finally, resistance of that one battery can be calculated subtracting resistances of other components from the sum:

\(\begin{aligned}{c}{R_2} &= {R_{t2}} - 3 \times 0.02 - 10\\{R_2} = 28 - 0.06 - 10\\{R_2} &= 18.06{\rm{ }}\Omega \end{aligned}\)

Therefore, internal resistance value is \({{\rm{R}}_{\rm{2}}}{\rm{ = 18}}{\rm{.06\Omega }}\).