Electromagnetism - High and Low-Pass Filters

The circuits below consist of two-element combinations of capacitors, diodes and resistors. V_in represents an ac-voltage with variable frequency. It is desired to build a circuit for which V_out ≈ V_in at high frequencies and V_out ≈ 0 at low frequencies. Which of the following circuits will perform this task?

(GR9677 #45)

Solution:

High-pass filter: V_out ≈ V_in at high frequencies, (ω → ∞)
Low-pass filter: V_out ≈ 0 at low frequencies (ω → 0)

Diode is a device to allow current flow in one direction only. A combination of Diode and Resistor will not affect the output frequency.
→ (B), (C) are FALSE

(A), (D), (E) circuits are combination of resistor and capacitor.

Capacitive reactance, X_C  = 1 / ωC
X_C → 0 for ω → ∞
X_C → ∞ for ω → 0

(A) FALSE
It is parallel combination of capacitor and resistor, voltage is uniform throughout the circuit.

(D) FALSE
It is Low-pass filter only.
Using Voltage Divider for series circuit: V_out = X_C / (R + X_C) V_in

X_C → 0 for ω → ∞

V_out = 0 / (R + 0) V_in ≈ 0

(E) TRUE
It is High and Low-pass filter.
Using Voltage Divider: V_out = R / (R + X_C) V_in

X_C → 0 for ω → ∞

V_out = R / (R + 0) V_in 
V_out ≈ V_in

X_C → ∞ for ω → 0
V_out = R / (R + ∞) V_in 
V_out ≈ 0

Answer: E

Notes: for similar problem see GR0177 #39

Electromagnetism - Resistor

In the circuit shown above, the resistances are given in ohms and the battery is assumed ideal with emf equal to 3.0 volts The resistor that dissipated the most power is

A. R₁
B. R₂
C. R₃
D. R₄
E. R₅

(GR9277 #32)

Solution:

Power, P = VI = I²R → P ∝ R

R₃ and R₄ are in parallel → R₃₄
R₃₄ and R₅  are in series → R₃₄₅
R₂ and R₃₄₅  are in parallel → R₂₃₄₅





R₁ is the highest resistance → dissipated the most power.

Answer: A  

Electromagnetism - Kirchoff's Law

See Question #32. The voltage across resistor R₄ is

A. 0.4 V
B. 0.6 V
C. 1.2 V
D. 1.5 V
E. 3.0 V

(GR9277 #33)

Solution:

Parallel: V_t = V₁ = V₂ = V₃ = ...

Series: V_t = V₁ + V₂ + V₃ + ...

V₄ = V₃ = V₃₄
V₃₄ + V₅ = V₂ 
V₂ + V₁ = V_t 

Or, using Kirchoff's Law:
Loop 1: V_t  = V₁ + V₂  
Loop 2: V_t  = V₁ + V₃₄ + V₅

From Question #32:
R₃₄ = 20
R₃₄₅ = 50
R₂₃₄₅ = 25

R_t = R₁ + R₂₃₄₅ = 50 + 25 = 75
I_t = V_t / R_t = 3/75 = I₁
V₁ =  I₁ R₁ = (3/75) × 50 = 2

Thus, V₃₄ + V₅ = V_t −V₁ = 3 − 2 = 1

V₃₄ must be less than 1
Only (A) and (B) could be TRUE

Since R₄ = R₅ but R₄ is in parallel with R₃ → V₄ must be less than V₅ 

Answer: A

Complete Calculation:

I₁ =  I₂ + I₃₄₅

R₃₄₅ = 50 = R₂
I₃₄₅ = I₂ =  I₁ / 2 = (3/75) / 2 = 3/150

R₃₄ and R₅ are in series
I₃₄₅ = I₃₄ = I₅
V₄ = V₃₄ = I₃₄ R₃₄ = (3/150) × 20 = 60/150 = 0.4

Electromagnetism - Electric Circuit

The battery in the diagram above is to be charged by the generator G. The generator has a terminal voltage of 120 volts when the charging current is 10 Amperes. The battery has an emf of 100 volts and an internal resistance of 1 Ohm. In order to charging current, the resistance R should be set at 

A. 0.1 Ω
B. 0.5 Ω
C. 1.0 Ω
D. 5.0 Ω
E. 10.0 Ω

(GR8677 # 24)

Solution:

Potential difference across the resistor R:

V = IR_t = V_Generator − V_Battery = 120 − 100 = 20 V

Total resistance:

R_t = R + R_internal = R + 1

V = IR_t = I(R + 1) = 10(R + 1) = 20 V
R = 1 Ω

Answer: C