Question

Find the equivalent resistance of the following circuit. If the battery voltage is 4.5V find the electric current of the circuit.

Answer 1

❏ Solution:-

From the above attachment,

➝The equivalent resistance between the terminal A&C is,

$\implies R_{AC}=(3+3) \:\Omega$

$\implies R_{AC}=6 \:\Omega$

➝The equivalent resistance between the terminal C&D is,

$\implies R_{CD}=\frac{2\times2}{2+2} \:\Omega$

$\implies R_{CD}=\frac{\cancel4}{\cancel4} \:\Omega$

$\implies R_{CD}=1 \:\Omega$

➝The equivalent resistance between the terminal D&E is,

$\implies R_{DE}=\frac{1\times1}{1+1} \:\Omega$

$\implies R_{DE}=\frac{\cancel1}{\cancel2} \:\Omega$

$\implies R_{DE}=0.5\:\Omega$

➝The equivalent resistance between the terminal A&G or the equivalent resistance of the whole circuit is ,

$\implies R= R_{AG}=R_{AC}+R_{CD}+R_{DE}$

$\implies R= R_{AG}=(6+1+0.5)\:\Omega$

$\implies R= R_{AG}=7.5\:\Omega$

$\implies\boxed{\large{\red{ R= 7.5\:\Omega}}}$

Hence, equivalent resistance(R) of the circuit is 7.5 ohms,

Now, Voltage(V) is = 4.5 volt,

$\therefore\textit{Current(I)through the circuit is} = \frac{V}{R}$

$\implies I = (\frac{4.5}{7.5})\:amp$

$\implies\boxed{\red{\large{ I =0.6\:amp}}}$

Hence, current(I) through the circuit is = 0.6 A.

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❏ Formula used:-

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✦ If the Resistances $R_1$,$R_2$,

$R_3$,........... are connected in series,

Then the equivalent resistance(R) will be ,

$\implies R=R_1+R_2+R_3+.......$

✦ If the Resistances $R_1$,$R_2$,

$R_3$,........... are connected in parallel,

Then the equivalent resistance(R) will be ,

$\implies \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_3}+\frac{1}{R_3}+..........$

N.B.

For Only $R_1$,$R_2$ ;

$\implies R=\frac{R_1\times R_2}{R_1+R_2}$

✦ If I amount of current passes through a resistance R of potential difference V then,

$\implies I=\frac{V}{R}$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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