Question

Draw a diagram showing three resistors with resistance 4 ohm, 12ohm and 20ohm in 1) series 2)parallel . Also calculate the effective resistance in each case....plz don't spam....​

Answer 1

Answer:

Effective resistance in :

1) Series = 36Ω

2)Parallel = 60/23 Ω

Explanation:

We know that,

Effective resistance in series = R series

$\boxed{R_{series}= R_{1} + R_{2} + R_{3}}$

Effective resistance in parallel = R parallel

$\boxed{\frac{1}{R_{parallel}} =\frac{1}{R_{1}}+\frac{1}{R_{2}} +\frac{1}{R_{3}} }$

Given that,

$R_{1}=4\Omega$

$R_{2}=12\Omega$

$R_{3}=20\Omega$

Putting these values in we get,

$R_{series} = 4+12+20 = 36\Omega$

$\frac{1}{R_{parallel}}= \frac{1}{{4}}+\frac{1}{{12}}+\frac{1}{{20}}$

$\frac{1}{R_{parallel}} = \frac{15+5+3}{60}= \frac{23}{60}$

${R_{parallel}}=\frac{60}{23}$Ω

Answer 2

Explanation:

${\red{\underline{\underline{\bold{Given:-}}}}}$

• Three resistors with resistance

1. 4Ω
2. 12Ω
3. 20Ω

${\blue{\underline{\underline{\bold{To\:Find:-}}}}}$

• The effective resistance

1. When resistors are connected in series
2. When resistors are connected in parallel

${\green{\underline{\underline{\bold{Solution:-}}}}}$

•$\bf R_{1} = 4Ω$

• $\bf R_{2} = 12Ω$

• $\bf R_{3} = 20Ω$

Case 1:-

Resistors connected in series:-

$\bf \: Effective \: \: resistance \:(R_{s}) = R_{1} + R_{2} + R_{3}$

$\bf \: =( 4 + 12 + 20)Ω$

$\bf \: =36Ω$

Case 2:-

Resistors connected in parallel:-

$\bf \: Effective \: \: resistance \:(R_{p} )$

$\bf \: \dfrac{1}{R_{p}} = \dfrac{1}{ R_{1}} + \dfrac{1}{ R_{2}} + \dfrac{1}{ R_{3}}$

$\bf = \dfrac{1}{4} + \dfrac{1}{12} + \dfrac{1}{20}$

$\bf = \dfrac{15 + 5 + 3}{60}$

$\bf \implies \dfrac{23}{60} = \dfrac{1}{ \: R_{p} }$

$\bf \implies \: R_{p} = \dfrac{60}{23} Ω$

$\boxed{ \bf \implies \: R_{s} = 36 Ω}$

$\boxed{ \bf \implies \: R_{p} = \dfrac{60}{23} Ω}$