Question

A student has two resistors- 2 Ω and 3 Ω. She has to put one of them in place of R₂ as shown in the circuit. The current that she needs in the entire circuit is exactly 9A. Show by calculation which of the two resistors she should choose.

Answer 1

Case 1:

She uses the 2 ohm resistor.

So, R₂ = 2 ohm

R₁ = 4 ohm

R₁ & R₂ are connected in parallel.

Let total resistance be R

\(\frac 1R =\frac 1{R_1} + \frac 1{R_2}\)

\(\frac 1{R_1} =\frac 1{4} + \frac 1{2}\)

\(\frac 1{R_1} =\frac {1 + 2}4 = \frac34\)

Therefore, R = \(\frac 43\)

Given, V = 12 volts

According to omh's law

\(V = \frac I R\)

So, \(I = \frac VR\)

\(= \frac{12}{(\frac 43)}\)

\(= \frac{12 \times 3} 4\)

\(= 9A\)

Case 2:

She uses the 3 ohm resistor.

So, R₂ = 3 ohm

R₁ = 4 ohm

R₁ & R₂ are connected in parallel.

\(\frac 1R =\frac 1{R_1} + \frac 1{R_2}\)

\(\frac 1{R_1} =\frac 1{4} + \frac 1{3}\)

\(\frac 1{R_1} =\frac {3+4}{12} = \frac7{12}\)

Therefore, R = \(\frac 7{12}\)

Given, V = 12 volts

According to omh's law

\(V = \frac I R\)

So, \(I = \frac VR\)

\(= \frac{12}{(\frac{12}7)}\)

\(= \frac{12 \times 7} {12}\)

\(= 7 A\)

The current she needs in the entire circuit is 9A.

Therefore, we choose case 1.

So, R₂ = 2 ohm

Answer 2

The overall current needed = 9A. The voltage is 12V 

Hence by Ohm’s Law V=IR, 

The resistance for the entire circuit = \(\frac{12}{9}\) = \(\frac{4}{3}\Omega\) = R 

R₁ and R₂ are in parallel. 

Hence, R=\(\frac{(R_1R_2)}{(R_1+R_2)}\)= \(\frac{4R_2}{(4+R_2)}\)

\(\because\, R\) = \(\frac{4}{3}\) 

\(\frac{4}{3}=\)\(\frac{4R_2}{(4+R_2)}\) 

\(16+4R_2=12R_2\\8R_2=16\)

R₂ = 2\(\Omega\)