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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).

6 years ago

R

R

R

R

Potential difference,V = 6V

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

Then,

1/R

1/R

1/R

or,R

Let total resistance be R

R

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

Now,

I

I

Similarly,I

I

and I

c). V across 2ohm resistor will be-

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

V = 1.5 * 2 = 3V

6 years ago

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

1/R

1/R

R

For total resistance,Rt = R

R

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

So I

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

V

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

5 years ago

lets find parallel resistance first

= 24/12 = 2ohmp = 1/4 + 1/6 + 1/12=12/24Rp1/R3 + 1/R2 + 1/R1= 1/Rp 1/R**current in circuit=i=V/R=6/4=1.5amp****V**_{s}=I(2)=(1.5)2=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**

total resistance=Rp + Rs = 2 + 2 = 4ohm

current flow is diff in all parallel resistors and potential diff is same in all parallel resistors but not same as series resistor.

V_{p} potential diff across parallel resistors,V_{s} potential diff aross series resistor

V=V_{p}+V_{s}

V_{p}=V-V_{s}=6-3=3V

11 months ago

Dear student,

Please find the attached solution to your problem.

The resultant resistance from the parallel arrangement

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

Hence, R_{p} = 2Ω

Now total resistance in circuit = R_{p} + R_{s} = 2 + 2 = 4Ω

a). current in circuit, I = V/R = 6/4 = 1.5 A

b). Now the potential difference across the circuit will be the sum of potential drop across the two arrangements.

Hence, V = V_{p} + V_{s}

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

V_{p} = V – V_{s} = 6 – 3 = 3V

current through R_{1}, I_{1} = V_{p}/R_{1} = 0.75 A

current through R_{2}, I_{2} = V_{p}/R_{2} = 3/6 = 0.5 A

current through R_{3}, I_{3} = V_{p}/R_{3} = 3/12 = 0.25 A

c). We have already calculated, V_{s} = 3 V

Thanks and regards,

Kushagra

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