Test: Internal Resistance & Cells in Series and in Parallel (December 5) - NEET MCQ

Given, ε = 15V,
r = 4Ω,
I = 2A
Now, for resistance of the resistors
ε − Ir = V = IR;
15 − 2 × 4=2 × R;
15 − 8  = 2R
R = 7/2 = 3.5Ω.
Terminal voltage of battery,
V = IR = 2 × 3.5 = 7V

In the parallel combination of three resistance's resistance is

or


= 2.02Ω

Between points A and B all resistances are combined in series
∴ R_eq = 3Ω + 4Ω + 5Ω + 6Ω
= 18Ω
Between points A and B all resistances are combined in series
∴ R_eq = 3Ω + 4Ω + 5Ω + 6Ω
= 18Ω

In each segment of the combination 3Ω and 2Ω resistance are connected in series separately.
∴ R' = +3 = 6Ω and R'' = 2 + 2 = 4Ω
R' and R'' are connected in parallel
∴ For first segment

 Similarly for second and third segment

Now segment is connected in series then the total resistance of combination is
R_eq = Re_q1 + R_eq2 + R_eq3
= 

Potential of 20V20V will be same across each resistance current




∴ Total current drawn from circuit

I = I₁ + I₂ + I₃

= 10 + 5 + 4 = 19A

To get maximum equivalent resistance all resistances must be connected in series
∴ (R_eq)_max = R + R + R...ntimes = nR
To get minimum equivalent resistance all resistances myst be connected in parallel.
∴ 

Here,
ε = 12V,  r = 3Ω, I = 1A, V = IR = ε − Ir
∴ 
= 12 − 3 = 9Ω
and V = IR = 1 × 9 = 9V

If one had considered a solid cylinder of radius b, one can suppose that it is made of two concentric cylinders of radius a and the outer part, joined along the length concentrically one inside the other.
If Ia and Ix are the currents flowing through the inner and outer cylinders
∵ I_total = I_b = I_a + I_x
⇒ VR_b = V/R_a + V/R_x
where R_b is the total resistance and R_x is the resistance of the tubular part.
∴  
But 
∴ 
∴ 

Wire of length 2π x 0.1 m of 12Ωm^-1 is bent to a circle.

Resistance of each part = 12 x π x 0.1
= 1.2π Ω
Total resistance = 0.6π Ω

For maximum equivalent or total resistance of the resistors must be combined in series.
R_eq = R₁ + R₂ = R₃ = 5 + 4.5 + 3
= 12.5Ω