Question

How will you connect three resistors of 4 Ω, 5 Ω and 6 Ω respectively so as to obtain a resultant resistance of 7.4 Ω? Draw the circuit diagram and show the calculation​

Answer 1

Given :

• Resistors = 4 Ω , 5 Ω and 6 Ω.

• Total internal resistance = 7.4 Ω.

To find :

• Connection of the three Resistors to get the equivalent resistance of 7.5 Ω.

• Circuit diagram of the circuit.

Solution :

Let us out the Resistors of 4 Ω and 6 Ω in parallel circuit and 5 Ω in series circuit.

First let us find the total resistance in the parallel circuit.

We know the formula for total resistance in a parallel circuit i.e,

$\boxed{\bf{\dfrac{1}{R_{e}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}} + ... + \dfrac{1}{R_{n}}}}$

Where :

• Rn = Total resistance.
• R = Resistors.

Now by using the Equation for total resistance in a parallel circuit and substituting the values in it, we get :

$:\implies \bf{\dfrac{1}{R_{e}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}}}$

$:\implies \bf{\dfrac{1}{R_{e}} = \dfrac{1}{4} + \dfrac{1}{6}}$

$:\implies \bf{\dfrac{1}{R_{e}} = \dfrac{3 + 2}{12}}$

$:\implies \bf{\dfrac{1}{R_{e}} = \dfrac{5}{12}}$

$:\implies \bf{R_{e} = \dfrac{12}{5}}$

$:\implies \bf{R_{e} = 2.4}$

$\boxed{\therefore \bf{R_{e} = 2.4\:\Omega}}$

Hence the total resistance in the parallel circuit is 2.4 Ω.

Now let us the find the total resistance in the series circuit :

We know the formula for total resistance in a series circuit i.e,

$\boxed{\bf{R_{e} = R_{1} + R_{2} + R_{3} + ... + R_{n}}}$

Where :

• Rn = Total resistance.
• R = Resistors.

$:\implies \bf{R_{e} = R_{1} + R_{2}}$

Here in our case , R2 will be total resistance in the parallel circuit .

$:\implies \bf{R_{e} = 5 + 2.4}$

$:\implies \bf{R_{e} = 7.4}$

$\boxed{\therefore \bf{R_{e} = 7.4\:\Omega}}$

Hence the total resistance in the series circuit is 7.4 Ω.

Since we have found the total cost by using all the Resistors , the equivalent resistance is 7.4 Ω.

Hence the resistors 4 Ω and 6 Ω are to be combined in the parallel circuit and resistor of 5 Ω to be connected in series.

Answer 2

Answer:

$\fbox{▪given}$

Three resistors of 4 Ω, 5 Ω and 6 Ω

respectively.

$\fbox{▪to \: find}$

➝ Arrange the given resistors in such a way that the resultant resistance is 7.5Ω

➝ circuit diagram for the arrangement of the given resistors.

$\fbox{▪solution}$

Arrange the resistors of value 4Ω and 6Ω in parallel arrangement and 5Ω resistors in series with 4Ω and 6Ω.

Formula for resistance for parallel arrangement-

$\red{\frac{1}{Req} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} + ...... + \frac{1}{Rn}}$

$\frac{ 1 }{Req} = \frac{1}{4} + \frac{1}{6} \\ \\➝ \frac{ 1 }{Req} = \frac{3 + 2}{12} = \frac{5}{12} \\ \\➝ Req = \frac{12}{5} Ω \\ ➝ \green{ Req = 2.4Ω}$

Formula for the arrangement of resistors in series-

$\pink{Req = R1 + R2 + R3..... + Rn}$

$Req = 2.4 + 5 \\ \bold \green{Req =7.4 Ω}$

For the circuit diagram refer to the above attachment.....

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