Question

find the total resistance and resistance value of R​

Answer 1

★ Concept:-

Here in this question, concept of ohm's law is used. We see that we are given the total current of circuit and potential difference of battery and there are 2 resistors connected in parallel of which value of 1 resistor is given and we have to find value of 2nd Resistor and total effective resistance in circuit. So firstly we will find the value of potential difference across 60Ω resistor. As potential difference divides in || connection of resistors, we can easily find the potential difference of 2nd Resistor by using the voltage of battery and 2nd voltage of 2nd resistor and further by applying ohms law we will solve the whole problem.

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★ Given that:-

• Current, I=0.3 A

• R₁=60Ω

• Battery voltage,V=6 Volts

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★ Solution:-

By applying ohms law for 60Ω resistor::

⇒ V₂=IR₂

By substituting values in formula::

⇒ V₂=(0.3A)×(60Ω)

⇒ V₂=18volts

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Now we know that voltage in parallel connection is expressed 3::

$\sf:\implies\dfrac{1}V}=\dfrac{1}{V_1}+\dfrac{1}{V_2}$

By substituting values::

$\sf:\implies\dfrac{1}6}=\dfrac{1}{18}+\dfrac{1}{V_1}$

$\sf:\implies\dfrac{1}6}-\dfrac{1}{18}=\dfrac{1}{V_1}$

Solving LHS by LCM::

$\sf:\implies\dfrac{3-1}{18}=\dfrac{1}{V_1}$

$\sf:\implies\dfrac{2}{18}=\dfrac{1}{V_1}$

Dividing by 2 from both Nr and Dn on LHS:;

$\sf:\implies\dfrac{2\div 2}{18\div 2}=\dfrac{1}{V_1}$

$\sf:\implies\dfrac{1}{9}=\dfrac{1}{V_1}$

After cross multiplication, we get::

$\boxed{\sf:\implies V_1=9\:volts}$

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Now applying ohm's law for 2nd resistor::

⇒ V₁=IR

By substituting values:

⇒ 9 V=(0.3 A)×(R)

Dividing both sides by 0.3::

⇒ 9 V÷0.3=(0.3 A)×(R)÷0.3

⇒ 30Ω= R

So the value of R is 30Ω.

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Now finding total effective resistance::

$\sf :\implies \dfrac{1}{R_p}=\dfrac{1}{R}+\dfrac{1}{R_2}$

By substituting values::

$\sf :\implies \dfrac{1}{R_p}=\dfrac{1}{30\Omega}+\dfrac{1}{60\Omega}$

$\sf :\implies \dfrac{1}{R_p}=\dfrac{2+1}{60\Omega}$

$\sf :\implies \dfrac{1}{R_p}=\dfrac{3}{60\Omega}$

$\sf :\implies \dfrac{1}{R_p}=\cancel{\dfrac{3}{60\Omega}}$

$\sf :\implies \dfrac{1}{R_p}=\dfrac{1}{20\Omega}$

By cross multiplication::

$\boxed{\sf :\implies R_p=20\Omega}$

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So the final answers are::

• R=30Ω

• Total resistance=20Ω

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