Answer: A.

84. What formula expresses correctly the voltage division rule application for k-th resistor if k of them are connected in series?

A. v_k=v_total*(R_eq/R_k) B.v_k=v_total*(R_k/R_eq) C. v_k=v_total*(R₁/R_eq) D. v_k=v_total/(R_k/R_eq)

E. v_k=v_total*(R_k-1/R_eq)

Solution:

 

The Voltage Division Rule for k-th resistor if k of them are connected in series is

Answer: B.

 

85. What is the equivalent resistance for the circuit below if all resistances are equal to R?

(between points A and D)

A.0.5R B.R C.1.5R D.2R E.3R

Solution: 1st way

 

 

The values of star sides will be three times less, that is R/3, so after reducing resistors in series there will be parallel combination of two resistors with values 4R/3 (R+R/3) in series with one R/3 resistor.

This gives 2R/3+R/3=R.

Answer: B.

D way

Points B and C are symmetrical respectively to points A and D, so potential of point B equals to potential of point C. Then we can ignore resistor between points B and C. As a result, we have got the circuit, where four resistors of R are connected in two groups of R/2 (two resistors of R in parallel). So resultant value of Req=R/2+R/2=R

86. What formula expresses correctly the current division rule application for k-th resistor if k of them are connected in parallel?

A. i_k=i_total*(R_eq/R_k) B.i_k=i_total*(R_k/R_eq) C. i_k=i_total*(R₁/R_eq) D. i_k=i_total*(G_k/G_eq)

E. i_k=i_total*(G_eq/G_k)

 

Solution:

The current division rule for k-th resistor if k of them are connected in parallel is