Question

A parallel combination of 3 resistance takes a current of 7.5A from a 30 volt battery.If the two resistors are of 10 ohm and 12 ohm .Find the third resistance?​

Answer 1

Given:

• Potential difference, V = 30V
• Current, I = 7.5 A
• $\sf r_1 = \bf{10 \Omega}$
• $\sf r_2 = \bf{12 \Omega}$

To find:

• Third resistance

SoluTion:

☯ Let the third resistor be $\bf r_3$.

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$\dag\;{\underline{\frak{Using\: ohm's\;law,}}}$

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$\star\;{\boxed{\sf{\purple{V = IR}}}}$

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$\dag\;{\underline{\frak{Putting\;values\;:}}}$

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$:\implies\sf 30 = 7.5 \times R$

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$:\implies\sf R = \dfrac{30}{7.5}$

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$:\implies{\underline{\boxed{\sf{\purple{R = 4 \Omega}}}}}\;\bigstar$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

⠀⠀⠀⠀☯ Here, R is the equivalent resistance And, Three resistors are in parallel combination.

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Therefore,

$\dag\;{\underline{\frak{We\;know\;that,}}}$

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$\star\;{\boxed{\sf{\pink{ \dfrac{1}{R} = \dfrac{1}{r_1} + \dfrac{1}{r_2} + \dfrac{1}{r_3}}}}}$

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$\dag\;{\underline{\frak{Putting\;values\;:}}}$

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$:\implies\sf \dfrac{1}{4} = \dfrac{1}{10} + \dfrac{1}{12} + \dfrac{1}{r_3}$

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$:\implies\sf \dfrac{1}{r_3} = \dfrac{1}{4} - \dfrac{1}{10} - \dfrac{1}{12}$

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$:\implies\sf \dfrac{1}{r_3} = \dfrac{15 - 6 - 5}{60}$

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$:\implies\sf \dfrac{1}{r_3} = \dfrac{15 - 11}{60}$

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$:\implies\sf \dfrac{1}{r_3} = \cancel{ \dfrac{4}{60}}$

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$:\implies\sf \dfrac{1}{r_3} = \dfrac{1}{15}$

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$:\implies{\underline{\boxed{\sf{\pink{r_3 = 15 \Omega}}}}}\;\bigstar$

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$\therefore$ Hence, The third resistor is of resistance $\bf 15 \Omega$.

Answer 2

Given ,

A parallel combination of 3 resistance takes a current of 7.5 A from a 30 volt battery and the two resistors are of 10 ohm and 12 ohm

Let , the third resistor of resistance be " R' "

We know that , the equivalent resistance in parrallel combination is given by

$\boxed {\tt{ \frac{1}{R} = \frac{1}{ R_{1} } + \frac{1}{ R_{2} } + .... + \frac{1}{ R_{n} } }}$

and the relationship between current and potential difference is

$\boxed{ \tt{V = IR}}$

Thus ,

The equivalent resistance will be

30 = 7.5 × R

R = 30/7.5

R = 4 ohm

Since , the three resistance are in parrallel combination

Thus ,

1/4 = 1/12 + 1/10 + 1/R'

1/R' = 1/4 - 1/12 - 1/10

1/R' = (120 - 40 - 48)/480

1/R' = 32/480

R' = 480/32

R' = 15 ohm

Therefore , the third resistor of resistance is 15 ohm

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