Question

How to find the specific resistance of the resistors using equivalent resistors in a parallel circuit?​

Answer 1

The resistors are connected across two points in the circuit in such a way that the current flowing through each resistor is the same and only one path is available for it to flow, then the resistors are said be connected in series.\In the figure above, three resistors R

1

,R

2

and R

3

are connected in series across the points A and B. Hence, the current (I) flowing through each of the resistors is same but the total voltage of the battery is divided.If the voltage drops across V

1

,V

2

and V

3

are R

1

,R

2

and R

3

respectively then,

V=V

1

+V

2

+V

3

…………….(i)

Now, if the resistor R instead of these three resistances R

1

,R

2

and R

3

is connected in such a way that the current flowing through the circuit remains the same, then R is called equivalent resistance of the circuit.For an equivalent resistor, V=IR…………….(ii)According to Ohm’s law,

R

1

voltage drops across, V

1

=IR

1

…………..(iii)

R

2

voltage drops across, V

2

=IR

2

…………..(iv)

R

3

voltage drops across, V

3

=IR

3

…………..(v)

Adding (iii), (iv) and (v):

V

1

+V

2

+V

3

=IR

1

+IR

2

+IR

3

……(vi)Or,

V=IR=I(R

1

+R

2

+R

3

)Or,

R=R

1

+R

2

+R

3

Thus, in series connection the equivalent resistor R is equal to the sum of all the resistors connected in series.

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Answer 2

Answer:

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Step-by-step explanation:

Imagine that you have a pipe that will permit a given flow rate (call it “ x “) at a certain starting water pressure.

Now, imagine that instead of one pipe, you use two pipes of that exact same size right next to (in parallel with) each other and using the exact same starting pressure.

What do you think that the new flow rate will be?

If you said, “2 • x”, you would be precisely correct.

Now, apply this identical idea using voltage, current, and resistors instead of pressure, flow, and friction losses in pipes.

If you have one resistor at a certain starting voltage giving a current of “ y “, and then you add a second identical resistor in parallel with the first, what will be your new current?

Now, if V = y • R for one resistor, it will be V = [ ? f( y ) ? ] • “the new R".

You now have 2 equations and 2 unknowns. So, solve for "the new R".

Hint:. [ ? f( y ) ? ] is an expression with y in it!

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