Answer: E. 26. Equivalent resistance of parallel arrangement is equal to 6 Ω

26. Equivalent resistance of parallel arrangement is equal to 6 Ω. Two resistors are connected in a parallel, one of them is 10 Ω. What is the second one?

A.15 Ω; B.20 Ω C. 10 Ω D. 12 Ω E. 8 Ω

 

Solution:

R_EQ for two parallel branches is . We know that R_EQ=6W and one of the resistors equals 10W. Let the second one will be XW. Substituting these values gives equation . Solving for X gives us X=15W.

Answer: A.

 

27. The circuit consists of a 10 Ω resistor in series with a parallel combination of two resistors with value of 15 Ω and 5 Ω. Current through 5 Ω resistor equals 6A. What total power absorbs by the circuit?

 

A.500W B.880W C. 560W D.740W E. 960W

 

Solution:

Current in lower branch is three times less than in upper one,

it equals 6/3=2A. P=I²R

P_upper=I²R=36*5=180W

P_lower= I²R=4*15=60W

P_mid=(6+2)²*10=640W

P_total=180+60+640=880W

Answer: B.

 

28. When ideal switch is closed, the voltage is equal to______ and current is___, but when it is open current is equal to _____ and voltage is _________.

 

A. any value, 0, any value,0 B. any value, 0, 0, any value

C. any value, any value, 0, 0 D. 0, any value, 0, any value

E. 0, any value, any value, 0

 

Solution:

When switch is closed, it’s a short circuit: v=0 when R=0 and i=any value.

When switch is opened it’s an open circuit: i=0 when R=∞ and v=any value.

Answer: D.

29. KVL states that

 

A. The sum of all voltages is 0. B. The sum of all voltages around the loop is 0.

C. The algebraic sum of all voltages around the loop is 0.

D. The algebraic sum of currents in a loop is 0.