Question

please take my question​

Answer 1

» Required Answer :

As in the given circuit diagram, we can see that $\sf { 12.0 Ω \: (R_1)}$ and $\sf { 6.00 Ω \: (R_2)}$ ;$\sf { 4.00 Ω \: (R_4)}$ and$\sf { 8.00 Ω \: (R_5)}$ are in parallel combination.

Therfore, by the formula :

• $\boxed { \large \bf { \dfrac{1}{R_p} = \dfrac{1}{R_1} +\dfrac{1}{R_2}...... }}$

Substituting values-

${ \large \bf { \dfrac{1}{R_p} = \dfrac{1}{R_1} +\dfrac{1}{R_2}}}$

$\bf {\implies \dfrac{1}{R_p} = \dfrac{1}{12} +\dfrac{1}{6}}$

$\bf {\implies \dfrac{1}{R_p} = \dfrac{1+2}{12}}$

$\bf {\implies \dfrac{1}{R_p} = \dfrac{3}{12}}$

$\bf {\implies R_p= \cancel{ \dfrac{12}{3}} = 4.00Ω}$

And in the second parallel combination:

${ \large \bf { \dfrac{1}{R_p} = \dfrac{1}{R_4} +\dfrac{1}{R_5}}}$

$\bf {\implies \dfrac{1}{R_p} = \dfrac{1}{4} +\dfrac{1}{8}}$

$\bf {\implies \dfrac{1}{R_p} = \dfrac{2+1}{8} }$

$\bf {\implies \dfrac{1}{R_p} = \dfrac{3}{8} }$

$\bf {\implies R_p= \cancel{ \dfrac{8}{3}} = 2.66Ω}$

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Now,the circuit will become in series combination.

By the formula:

• $\boxed { \large \bf { R_s = R_1 + R_2...... }}$

Substituting values, we get-

${ \large \bf { R_s = R_1 + R_2 + R_3}}$

$\bf { \implies 4.00Ω + 5.00 Ω + 2.66Ω}$

$\bf { \implies 9.00Ω+ 2.66Ω}$

$\bf \red { \implies 11.66Ω}$

So, we got the answer. The net resistance is $\bf { 11.66Ω}$.

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More Information:

• The property of a conductor due to which it opposes the flow of current through it is called resistance.
• The resistance of a conductor depends on the length, thickness, nature of material and temperature of the conductor.
• The SI unit of resistance is ohm (Ω).

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