Question

Two resistors of resistance 12 ohm and 6 ohm are connected in parallel to a battery of 12 V.
a) Calculate the equivalent resistance of the network. b) obtain the current in 12 ohm and 6 ohm resistors​

Answer 1

Explanation:

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

• Two resistors of 12Ω and 6Ω are connected in parallel.
• Voltage of potential difference = 12V

We are asked to calculate equivalent resistance of the network and the current in 12Ω and 6Ω resistors.

★ Calculating the equivalent resistance of the network :

Since, the resistors are connected in parallel combination, so we need to apply the formula of parallel combination of resistances.

Whenever the resistors are connected in parallel combination, then equivalent resistance is given by,

$\\ \longrightarrow \quad\pmb{\boxed{ \sf { \dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots \dfrac{1}{R_n} }}}$

Here,

• $\sf R_1$ = 6Ω
• $\sf R_2$ = 12Ω

Substituting values to find the equivalent resistance of the network,

$\\ \longrightarrow \quad \sf { \dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} }$

$\\ \longrightarrow \quad \sf { \dfrac{1}{R_P} = \dfrac{1}{12} + \dfrac{1}{6} }$

$\\ \longrightarrow \quad \sf { \dfrac{1}{R_P} = \dfrac{1 + 2}{12} }$

$\\ \longrightarrow \quad \sf { \dfrac{1}{R_P} = \dfrac{3}{12} }$

On reciprocating both sides,

$\\ \longrightarrow \quad \sf { R_P = \dfrac{12}{3} }$

$\\ \longrightarrow \quad \bf \underline { R_P = 4 \; \Omega }$

Therefore, the equivalent resistance of the network is 4Ω.

★ Obtaining the current in 12Ω and 6Ω resistors :

As we know that,

$\\ \longrightarrow \quad\pmb{\boxed{ \sf {V = I R}}}$

• V denotes potential difference
• I denotes current
• R denotes resistance

Finding current flowing in 12Ω resistor :

$\\ \longrightarrow \quad\pmb{\boxed{ \sf {V = I R}}}$

• V = 12V
• I = ?
• R = 12Ω

$\\ \longrightarrow \quad \sf { 12 = 12(I)}$

$\\ \longrightarrow \quad \sf { \dfrac{12 }{12} \; A= I}$

$\\ \longrightarrow \quad \bf \underline { 1\; A= I}$

Therefore, current in 12Ω resistor is 1 Ampere.

Finding current flowing in 6Ω resistor :

$\\ \longrightarrow \quad\pmb{\boxed{ \sf {V = I R}}}$

• V = 12V
• I = ?
• R = 6Ω

$\\ \longrightarrow \quad \sf { 12 = 6(I)}$

$\\ \longrightarrow \quad \sf { \dfrac{12 }{6} \; A= I}$

$\\ \longrightarrow \quad \bf \underline { 2 \; A= I}$

Therefore, current in 6Ω resistor is 2 Ampere.

Answer 2

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