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A child’s electronic toy is supplied by three \(1.58{\rm{ }}V\) alkaline cells having internal resistances of \({\bf{0}}.{\bf{0}}2{\bf{00}}{\rm{ }}{\bf{\Omega }}\) in series with a \(1.53{\rm{ }}V\) carbon-zinc dry cell having a \({\bf{0}}.1{\bf{00}}{\rm{ }}{\bf{\Omega }}\) internal resistance. The load resistance is \(1{\bf{0}}{\rm{ }}{\bf{\Omega }}\).

(a) Draw a circuit diagram of the toy and its batteries.

(b) What current flows?

(c) How much power is supplied to the load?

(d) What is the internal resistance of the dry cell if it goes bad, resulting in only 0.500 W being supplied to the load?

Short Answer

Expert verified

a) Circuit diagram of the toy and it’s batteries is

b) Value of current flow through is\(I = 4.89\;A\)

c) Value of power supplied to the electric toy is\(P = 239\;W\).

d) Internal resistance of the dry cell is \({R_2} = 18.06{\rm{ }}\Omega \).

Step by step solution

01

Definition of loop rule in circuits.

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.

02

Information Provided

  • 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: \(10{\rm{ }}\Omega \)
03

Circuit diagram and Calculation of current in circuit using the loop rule.

a)

Circuit diagram of the toy and its batteries is

04

Calculation of current flows

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

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

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

05

Calculation of power supplied to the electric toy.

c)

Power supplied is equal to:

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

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

06

Calculation of internal resistance.

d)

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

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

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

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

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

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