Question

10 ohm, 15 ohm resistors are connected ina parallel to a battery of 12V. What is the effective Resistance in the circuit

Answer 1

Answer:

6Ω

Explanation:

As per the provided information in the given question, we have :

• 10Ω & 15Ω resistors are connected in parallel combination to a battery of 12V.

We are asked to calculate the effective resistance of the circuit. Let,

• $\sf R_1$ = 10Ω
• $\sf R_2$ = 15Ω

As these tow resistors have been connected in the parallel combination, so equivalent resistance will be given by,

$\longrightarrow \sf{\quad { \dfrac{1}{R_{(1,2)}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} }}$

Substituting the values of the resistors.

$\longrightarrow \sf{\quad { \dfrac{1}{R_{(1,2)}} =\Bigg \{ \dfrac{1}{10} + \dfrac{1}{15} \Bigg \} \; \Omega}}$

Taking the LCM in the RHS and simplifying further.

$\longrightarrow \sf{\quad { \dfrac{1}{R_{(1,2)}} =\Bigg \{ \dfrac{3 + 2 }{30} \Bigg \} \; \Omega}}$

Performing addition in the numerator of the fraction in RHS.

$\longrightarrow \sf{\quad { \dfrac{1}{R_{(1,2)}} =\Bigg \{ \dfrac{5 }{30} \Bigg \} \; \Omega}}$

Reciprocating both sides.

$\longrightarrow \sf{\quad { R_{(1,2)} = \dfrac{30}{5} \; \Omega}}$

Dividing 30 by 5.

$\longrightarrow \quad \underline{\boxed{ \textbf{\textsf{R}}_{\textbf{\textsf{(1,2)}}} = \textbf{\textsf{6}} \;\textbf{\textsf{\Omega}}}}$

Therefore, effective resistance of the circuit is 6Ω.

$\underline{ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}$

Points to remember :

When resistors are connected in parallel combination then effective resistance of the circuit is given by,

$\longrightarrow \sf{\quad { \dfrac{1}{R_{P}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots \dfrac{1}{R_n}}}$

When resistors are connected in series combination then effective resistance of the circuit is given by,

$\longrightarrow \sf{\quad { R_S = R_1 + R_2 + \dots R_n}}$

Answer 2

Answer:

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