Question

Q.8. Two resistance of 10 and 20 are connected i) in series and ii) in
parallel, in turn to a 9V battery. Calculate the ratio of the power
consumed in the combination of resistors in each case.​

Answer 1

Answer:

4.5

Explanation:

For case (i) resistances are in series , $R_s$ = (10+20) $\Omega$ = 30$\Omega$

$P_s = V^2\times R_s$ = 9²× 30= 2430 Watt

For case (ii) resistances are in parallel , $R_p = (10^{-1}+20^{-1})^{-1}\Omega= 6.67 \Omega$

$P_p = V^2\times R_p$ = 9²× 6.67=  540 Watt

ratio of power consumed = $\frac{P_s}{P_p} = \frac{2430}{540}$  = 4.5

Answer 2

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✤ Required Answer:

✒ GiveN:

• Resistors are 10Ω and 20Ω
• They are connected in series
• Then, they are also connected in parallel.
• Voltage of battery = 9V

✒ To Find:

• Ratio of the Combination of resistors in each case....?

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✤ How to Solve?

When we have voltage and resistance, we can find the power of combination of resistance by:

• P = V² / Req.

And, we know that formula for finding the equivalent resistance in series and parallel combination:

Series:

$\large{ \boxed{ \rm{Req. = R1 + R2 + R3....}}}$

Parallel:

$\large{ \boxed{ \rm{Req. = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} .....}}}$

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✤ Solution:

⚡1st case:

We have,

• Resistors = 10Ω and 20Ω

Equivalent resistance,

➝ Req. = 10Ω + 20Ω

➝ Req. = 30Ω

Finding the power of the combination,

➝ P = V² / R

➝ P = 9² / 30 watt

➝ P = 27 / 10 watt

➝ P = 2.7 watt

⚡ 2nd case:

Equivalent resistance,

➝ 1/Req. = 1/10Ω + 1/20Ω

➝ 1/Req. = 3/20 Ω

➝ Req. = 20/3 Ω

Finding the power of the combination,

➝ P = V² / R

➝ P = 9² / 20/3 watt

➝ P = 9² × 3 / 20 watt

➝ P = 12.15 watt

❍ Ratio of the power in each case:

➝ Ps / Pp = 2.7 watt / 12.15 watt

➝ Ps / Pp = 0.2223 (approx)

⚘ Hence, solved !!

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