A child's electronic toy is supplied by three 1.58 V alkaline cells having internal resistances...

Question:

A child's electronic toy is supplied by three 1.58 V alkaline cells having internal resistances of 0.0230{eq}\Omega {/eq} in series with a 1.53 V carbon-zinc dry cell having a 0.170 internal resistance. The load resistance is 11.0{eq}\Omega {/eq}.

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

(b) What current (in A) flows?

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

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

Battery Connection:

If a circuit contains more than one DC battery, then all the batteries should be connected in series. By a series connection of batteries, maximum battery voltages can be obtained. A battery also has some internal losses. These losses are shown by the battery's internal resistance.

Part (a);

Let the internal resistance of the alkaline cell be {eq}r_1 {/eq} and carbon-zinc dry cell be {eq}r_2 {/eq}, the load resistance be R . We know that all the voltage sources should be connected in series to obtain a higher source voltage. The circuit diagram of the toy circuit can be shown as,

Part (b);

Let the current in the circuit be I.

We can get the circuit current by applying Kirchoff's voltage law as follows,

{eq}\begin{aligned} -I{{r}_{1}}+1.58\text{ V}-I{{r}_{1}}+1.58\text{ V}-I{{r}_{1}}+1.58\text{ V}-I{{r}_{2}}+1.53\text{ V}-IR&=0 \\ -I\left( 0.023\text{ }\Omega \right)+1.58\text{ V}-I\left( 0.023\text{ }\Omega \right)+1.58\text{ V}-I\left( 0.023\text{ }\Omega \right)+1.58\text{ V}-I\left( 0.17\text{ }\Omega \right)+1.53\text{ V}-I\left( 11\text{ }\Omega \right)&=0 \\ 6.27\text{ V}-11.239I&=0 \\ I&=\frac{6.27}{11.239} \\ &=0.56\text{ A} \\ \end{aligned} {/eq}

Hence, the circuit current is {eq}0.56\text{ A} {/eq}.

Part (c);

The power supply in the load can be determined as,

{eq}\begin{aligned} P&={{I}^{2}}R \\ &={{\left( 0.56\text{ A} \right)}^{2}}\times 11\text{ }\Omega \\ &=3.45\text{ W} \\ \end{aligned} {/eq}

Hence, the power consumed in the load is {eq}3.45\text{ W} {/eq}.

Part (d);

Let the new current in the circuit be {eq}I_n {/eq}.

For new power supply, the current in the circuit is given by using the electric power formula as,

{eq}\begin{aligned} {{P}_{n}}&=I_{n}^{2}R \\ {{I}_{n}}&=\sqrt{\frac{{{P}_{n}}}{R}} \\ &=\sqrt{\frac{0.5\text{ W}}{11\text{ }\Omega }} \\ &=0.21\text{ A} \\ \end{aligned} {/eq}

Again use KVL in the circuit, for new current in {eq}I_n {/eq} to get the internal resistance of dry cell {eq}r_2 {/eq}.

{eq}\begin{aligned} -{{I}_{n}}{{r}_{1}}+1.58\text{ V}-{{I}_{n}}{{r}_{1}}+1.58\text{ V}-{{I}_{n}}{{r}_{1}}+1.58\text{ V}-{{I}_{n}}{{r}_{2}}+1.53\text{ V}-{{I}_{n}}R&=0 \\ -\left( 0.21\text{ A} \right)\left( 0.023\text{ }\Omega \right)+1.58\text{ V}-\left( 0.21\text{ A} \right)\left( 0.023\text{ }\Omega \right)+1.58\text{ V}-\left( 0.21\text{ A} \right)\left( 0.023\text{ }\Omega \right)+1.58\text{ V}-\left( 0.21\text{ A} \right){{r}_{2}}+1.53\text{ V}-\left( 0.21\text{ A} \right)\left( 11\text{ }\Omega \right)&=0 \\ 3.95-0.21{{r}_{2}}&=0 \\ {{r}_{2}}&=\frac{3.95}{0.21} \\ &=18.80\text{ }\Omega \\ \end{aligned} {/eq}

Hence, the new internal resistance is {eq}18.80\text{ }\Omega {/eq}.

Electrical Circuits Lesson for Kids

from

Chapter 2 / Lesson 1
19K

An electrical circuit is created when a power source generates enough force to move electrons through a conductor to an electrical appliance. Discover how closed and open circuits and on-off switches regulate the flow of electricity to an electrical appliance like a light bulb.