Test: Combination of Resistors: Series & Parallel (NCERT) - NEET MCQ

In series combination, current across its circuit components is always constant and in parallel combination the voltage across the circuit components is constant.

• In a parallel combination of resistances, the voltage drop across all the resistors is the same and therefore, the current across the resistances depends on the value of resistance. If the resistance is small, the current flowing through it will be more.
• In a series combination of the resistances, the current through all the resistors is constant whereas the voltage differs for the resistors.
• Similarly, if the same voltage is applied across all the resistors, the power is inversely proportional to the resistance. So, greater the resistance, smaller will be the power and if the current is the same in the circuit, the power will be greater for the higher resistance.
  So, A → s
  B → r
  C → p
  D → q

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

In the parallel combination of three resistance's resistance is

or


= 2.20Ω

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

Three resistors of 3Ω each in parallel will give


R_eq = 3 + 1 + 3 = 7Ω.

Let Resistance of each resistor is = R
In series connection:
R_eq1 = 4R = S
R = S/4 → (1)
In parallel connection
R_eq2 = R4 = P
R = 4/P → (2)
Equating (1) and (2)
S/4 = 4P
S = 16P
n = 16

According to the given circuit 10Ω and 10Ω resistances are connected in series.
∴ R′ = 10 + 10 = 20Ω
Again 10Ω and 10Ω resistances are connected in series
∴ R′′ = 10 + 10 = 20Ω
R′, R′′ and 10Ω all connected in parallel than
∴ 
= 

R_eq = 5Ω

According to option (d) 2Ωand3Ω are combined in a parallel combination. Hence equivalent resistance

R' combined in series to 1Ω
∴ R_eq = 
Hence, the combination scheme in option (d) is correct.

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
= 

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.
∴ 

For a equivalent resistance betweeen A and B.
5Ω and 8Ω resistance are connected in series. R' their equivalent resistance is parallel to 6Ω
∴ R = 5 + 8 = 13Ω and 
R" = 78/19
Now 4Ω, R" an d 5Ω resistances are connected in series equivalent resistance between ,A and B
∴ 

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

Resistance per unit length of ring,

Length of sections ADB and ACB are rθ and r(2π − θ)
∴ Resistance of section ADB,

and resistance of section ACB,
R₂ = ρr(2π − θ)
= 
Now, R and R2 are connected in parallel between A and B then

= 
Putting θ = 45^∘ = π/4 rad and R = 15Ω

= 
= 1.64Ω

Resistance of a wire in terms of connectivity (σ) is given by

where l is the length and A is area of cross section of wire respectively.
∴ Rs = R1 + R2
⇒ 
where σ_s is the effective conductivity 

σ_{s =} 2σ₁σ₂ / σ₁  + σ₂

Potential of 20V20V will be same across each resistance current




∴ Total current drawn from circuit

I = I₁ + I₂ + I₃

= 10 + 5 + 4 = 19A

Since the voltage across the circuit is constant Then current through 3Ω resistor

The current through 4 £2 resistor 

and the current through 5 £2 resistor,

The equivalent circuit of the given circuit will be reduced to as shown in figure.

Total resistance of the circuit
= 
= 
Total current in the circuit = 
Reading of ammeter =  
= 2.18A

Let current in the circuit is I. Then total resistance in the circuit.
R = R₁ + R₂ + R₃ = 2 + 4 + 5 = 11Ω
∴ 
The potential drop across 2Ω resistance

The potential drop across 4Ω resistance

The potential drop across 5Ω resistance

Hence(V₁, V₂, V₃) = (2.18, 4.36, 5.45)V