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Question: -In the given circuit the current flowing through the resistance is \(20\;{\rm{ohms}}\) is \(0.3\;{\...

-In the given circuit the current flowing through the resistance is 20  ohms20\;{\rm{ohms}} is 0.3  ampere0.3\;{\rm{ampere}} while the ammeter reads 0.8  ampere0.8\;{\rm{ampere}}. What is the value of R1{R_1} in ohms?

Explanation

Solution

Here, first we analyze the circuit and then approaches to next step. The voltage through the 20  ohms20\;{\rm{ohms}} resistance is equal to the voltage across the circuit. It gives us the equivalent resistance of the circuit. We can see that the all three resistances are in parallel with each other. Then we will use the formula for resistances in parallel to find the value of resistance R1{R_1}.

Complete step by step answer:
Given: We have three resistances in the given circuit. The resistances are R2=20  ohms{R_2} = 20\;{\rm{ohms}} , R3=15  ohms{R_3} = 15\;{\rm{ohms}} and R1{R_1}. The current flowing through the resistance of R2=20  ohms{R_2} = 20\;{\rm{ohms}} is I2=0.3  ampere{I_2} = 0.3\;{\rm{ampere}}. The total current in the ammeter is I=0.8  ampereI = 0.8\;{\rm{ampere}}.
The current through the ammeter is given, the potential across the ammeter will be equal to the voltage across R2=20  ohms{R_2} = 20\;{\rm{ohms}} resistance because all three resistance are in parallel with each other.
V2=V{V_2} = V
Here, V is the potential across the ammeter and V2{V_2} is the potential across the resistance of 20 ohms.
I2R2=ReqI{I_2}{R_2} = {R_{eq}}I
Here, Req{R_{eq}} is the equivalent resistance of the circuit.
We substitute the values in above relation,
0.3×20=Req×0.8     6=Req×0.8     Req=60.8     Req=152  ohms 0.3 \times 20 = {R_{eq}} \times 0.8\\\ \implies 6 = {R_{eq}} \times 0.8\\\ \implies {R_{eq}} = \dfrac{6}{{0.8}}\\\ \implies {R_{eq}} = \dfrac{{15}}{2}\;{\rm{ohms}}
We write the formula for the resistance in parallel to find the resistance R1{R_1}.
1Req=1R1+1R2+1R3\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}}
Now, we substitute the values in above relation,
215=1R1+120+115     1R1=83460     1R1=160 \dfrac{2}{{15}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{20}} + \dfrac{1}{{15}}\\\ \implies \dfrac{1}{{{R_1}}} = \dfrac{{8 - 3 - 4}}{{60}}\\\ \implies \dfrac{1}{{{R_1}}} = \dfrac{1}{{60}}
We perform cross multiplication and multiply R1{R_1} to 1 and 60 to 1. It gives us the value of resistance R1{R_1}.
Then, we write the obtained value of R1{R_1}.
R1=60  ohms{R_1} = 60\;{\rm{ohms}}

Therefore, the value of resistance R1{R_1} is 60  ohms60\;{\rm{ohms}}.

Note:
In this question, the students must have the knowledge of potential difference and knowledge of the term diode. This question can alternatively be solved by Kirchhoff's voltage law. Kirchhoff’s gives two laws, first current law and voltage law. Kirchhoff’s current law states that the summation of all the current in the circuit is equal to zero. And Kirchhoff’s voltage law states that the summation of all the voltage in the circuit is equal to zero.