# 3 resistors 4 ohm,6 ohm and 12 ohm are connected in parallel.The combination of the above resistors is connected in series to a resistance of 2 ohm and then to a battery of 6 volt. draw a circuit diagram and calculate: a)current in the main circuit. b)current flowing through each of the resistors in parallel. c)potential difference across the 2 ohm resistor. (diagram is not necessary).

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3 resistors 4 ohm,6 ohm and 12 ohm are connected in parallel.The combination of the above resistors is connected in series to a resistance of 2 ohm and then to a battery of 6 volt. draw a circuit diagram and calculate:a)current in the main circuit.b)current flowing through each of the resistors in parallel.c)potential difference across the 2 ohm resistor.(diagram is not necessary).

## 5 Answers

**Given-**R

_{1}= 4ohm

R

_{2 }= 6ohm

R

_{3}= 12ohm

R

_{s}= 2ohm (Series resistor)

Potential difference,V = 6V

Let the resistance of all three resistors which are connected in parallel be R

_{p}

Then,

1/R

_{p }= 1/R

_{1}+1/R

_{2}+ 1/R

_{3}

1/R

_{p }= 1/4 + 1/6 + 1/12

1/R

_{p}= 6/12

or,R

_{p}= 12/6 = 2ohm

Let total resistance be R

_{t }= R

_{p}+R

_{s}

R

_{t =}2 + 2 = 4ohm

a).So current,I in the main circuit will be V=IR

therefore, I = V/R = 6/4 = 1.5A which is total current.

b) Since in parallel the current will divided into three resistors connected in parallel.Let the current across all the three be I

_{1},I

_{2}and I

_{3}.

Now,

I

_{1 }= V/R

_{1}(Since potential difference is same as they are connected in parallel)

I

_{1 }= 6/4 = 1.5A

Similarly,I

_{2}= V/R

_{2}

I

_{2}= 6/6 = 1A

and I

_{3 }= 6/12 = 0.5A

c). V across 2ohm resistor will be-

V = I*R ( I will be total current)

V = 1.5 * 2 = 3V

_{s}).Now,

a).The first part of question is asking about the current in the main circuit.So first we have to find the total resistance which we suppose to designated as R

_{t}.For parallel combination

1/R

_{p }= 1/R

_{1}+ 1/R

_{2}+ 1/R

_{3 }

1/R

_{p}= 1/4 + 1/6 + 1/12

R

_{p}= 12/6 = 2ohm

For total resistance,Rt = R

_{p}+ R

_{s}(Since all three connected in series with the 2ohm resistor)

R

_{t}= 2 + 2 = 4ohm

According to Ohm’s law,

V = I*R

I = V/R = 6/4 = 1.5A (Since the whole combination is connected to a battery of 6V)

b).Second part is asking the current through each resistor connected in parallel.Since in parallel combination current will get divided into three part each through the three resistors(I

_{1},I

_{2}and I

_{3}).Voltage in such combination remains same.

So I

_{1}= V/R

_{1}, I

_{2}= V/R

_{2}and I

_{3 }= V/R

_{3}which we calculated as explained above.

3).Third part is asking the voltage across 2ohm resistor.Let it be V

_{s}

V

_{s}= I * R

_{s }

For that main current which is calculated as above will be used as this resistor is connected in series so the main current will passes through it also.

V

_{s}= 1.5 * 2 = 3V

**current in circuit=i=V/R=6/4=1.5amp**

_{p}potential diff across parallel resistors,V

_{s}potential diff aross series resistor

_{p}+V

_{s}

**V**

_{s}=I(2)=(1.5)2=3V_{p}=V-V

_{s}=6-3=3V

**current through R**

_{1 }is I_{1}=V_{p}/R_{1}=3/4 amp**current through R**

_{2 }is I_{2}=V_{p}/R_{2}=3/6=1/2amp**current through R**

_{3 }is I_{3}=V_{p}/R_{3}=3/12=1/4amp**approve if it is r8**

_{p}= 1/4+ 1/6 + 1/12 = 6/12

_{p}= 2Ω

_{p}+ R

_{s}= 2 + 2 = 4Ω

_{p}+ V

_{s}

_{s}= I(2) = (1.5)2 = 3V

_{p}= V – V

_{s}= 6 – 3 = 3V

_{1}, I

_{1}= V

_{p}/R

_{1}= 0.75 A

_{2}, I

_{2}= V

_{p}/R

_{2}= 3/6 = 0.5 A

_{3}, I

_{3}= V

_{p}/R

_{3}= 3/12 = 0.25 A

_{s}= 3 V