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

Electromagnetism - Impedance

An alternating current electrical generator has a fixed internal impedance R_g + jX_g and is used to supply power to a passive load that has an impedance R_g + jX₁, where j =√(−1). R_g ≠ 0 and X_g ≠ 0. For a maximum power transfer between the generator and the load, X₁ should be equal to

A. 0
B. X_g
C. −X_g
D. R_g
E. −R_g

(GR8677 #64)

Solution:

Maximum power transfer theorem (Impedance matching): 

Maximum power is transferred from a source to a load when

• Load resistance = the internal resistance of the source, or 
• Load impedance = complex conjugate source impedance (Z_l = Z_s^*)

Z_s = R_g + jX_g  with  j =√(−1)

Z_s^* = R_g − jX_g

For max power transferred: 
Z_l  = Z_s^*
X₁ = −X_g

Answer: C