Figure shows two cylindrical uniform specimen of electrical resistances with their characteristics. Resistivity : ρ1 = ρ ρ2 = 2ρ Length : ℓ1 = ℓ ℓ2 = 2ℓ Area of corss-section : A1 = A A2 = A/2 No. of electron per unit volume : n1 = n n2 = 2n R1 and R2 are connected in circuit as shown in figure. Voltmeter and ammeter connected in the circuit are ideal. Comparison of electric field, drift speed and power disipated per unit volume should be done only if R1 carries current. Column-I Column-II (A) S1 is open S2 is closed (P) Reading of voltmeter is maximum (B) S1 is closed S2 is open (Q) Reading of ammeter is maximum (C) S1 and S2 both are closed (R) Magnitude electric field in R2 is greater than in R1. (D) S1 and S2 both are open (S) Drift speed of electrons is greater in R2 than in R1 (T) Power dissipated per unit volume is greater in R2 than in R1.

Engineering

Physics

Ohm Law and Current Basics

Question

Figure shows two cylindrical uniform specimen of electrical resistances with their characteristics.

Resistivity :	ρ₁ = ρ	ρ₂ = 2ρ
Length :	ℓ₁ = ℓ	ℓ₂ = 2ℓ
Area of corss-section :	A₁ = A	A₂ = A/2
No. of electron per unit volume :	n₁ = n	n₂ = 2n

R₁ and R₂ are connected in circuit as shown in figure. Voltmeter and ammeter connected in the circuit are ideal. Comparison of electric field, drift speed and power disipated per unit volume should be done only if R₁carries current.

Column-I	Column-II
(A) S₁ is open S₂ is closed	(P) Reading of voltmeter is maximum
(B) S₁ is closed S₂ is open	(Q) Reading of ammeter is maximum
(C) S₁ and S₂ both are closed	(R) Magnitude electric field in R₂ is greater than in R₁.
(D) S₁ and S₂ both are open	(S) Drift speed of electrons is greater in R₂ than in R₁
	(T) Power dissipated per unit volume is greater in R₂ than in R₁.

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$R_{1} = \frac{ρ 1}{A} = R (say) R_{2} = \frac{2 ρ \cdot 21}{\frac{A}{2}} = \frac{8 ρ 1}{A} = 8 R$

Case 1: S₁ is open, S₂ is closed
all resistance are active.
   Req = 5R
$i_{1} = \frac{ε}{5 R}$ (current through R₁) – it is also the reading of ammeter
$i_{2} = \frac{i_{1}}{2} = \frac{ε}{10 R}$ (current through R₂)
Voltmeter reading $= i_{2} R_{2} = \frac{ε}{10 R} \times 8 R = \frac{4 ε}{5}$
Current density in $R_{1}, J_{1} = \frac{i_{1}}{A}$
Current density in $R_{2}, J_{2} = \frac{\frac{l_{1}}{2}}{\frac{A}{2}} = \frac{i_{1}}{A} \Rightarrow J_{1} = J_{2}$
Drift speed, $V_{2} = \frac{J}{ne} \Rightarrow$ drift speed in R₁ is greater than drift speed in R₂
Electric field, E = J · P ⇒ E in greater in R₂ than in R₁.
Power dissipated per unit volume $= \frac{i^{2} R}{1 A} = \frac{i^{2} R}{A \cdot 1 A} = \frac{i^{2} P}{A^{2}} = {ρJ}^{2}$
⇒ Power dissipated per unit volume in greater in R₂ than in R₁.
Case II:   S₁ is closed and S₂ is open
       Only resistance in circuit is R₂ = 8R
Ammeter reading = $\frac{ε}{8 R}$
Voltmeter reading = ∈
comparison is meaningless.
Case III:   S₁ and S₂ bolt are closed

R₁ in by-passed.
$Req = \frac{R_{2}}{2} = 4 R$
   Ammeter reading = $\frac{ε}{4 R}$
   Voltmeter reading = ∈
   comparison is meaningless.
Case IV:   S₁ and S₂ bolt are open.
       Req = R + 8R = 9R
   Ammeter reading $= \frac{ε}{9 R}$
   Voltmeter reading = $\frac{8}{9} ε$
   Current density in $R_{1}, J_{1} = \frac{i}{A}$
   Current density in $R_{1}, J_{1} = \frac{i}{\frac{A}{2}} m = 2 J_{1}$
   Drift speed $= \frac{J}{ne} \Rightarrow$ Drift speed is same in R₁ and R₂
   Electric Field, E = Jr
⇒ electric field is greater in R₂ than in R₁
   Power dissipated per unit volume = rJ²
⇒ It is greater in R₂ than in R₁. Ans.

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