12. Consider a 75 W light bulb and 150 W bulb both designed to operate at 110 V.

Question

12. Consider a 75 W light bulb and 150 W bulb both designed to operate at 110 V. When both light bulbs are lit, which statement is correct?

A) The 75 W bulb has the same resistance as the 150 W bulb

B) The 75 W bulb has twice the resistance of the 150 W bulb

C) The 75 W bulb has 4 times the resistance of the 150 W bulb

D) The 75 W bulb has half of the resistance of the 150 W bulb

E) The 75 W bulb has one fourth of the resistance of the 150 W bulb

13. A copper wire of length L and cross sectional area A is used in an electrical application. If the same wire is cut in half and the two pieces are connected in parallel in the same application, how does the wire resistance in the latter connection compare to the wire resistance in the initial connection?

A) The resistance is the same since it is the same wire

B) The resistance is doubled

C) The resistance is quadrupled

D) The resistance is halved

E) The resistance is quartered

14. As resistors are added to series to a circuit, the current in the circuit ___

A) decreases

B) increases

C) first decreases, then increases

D) first increases, then decreases

E) remains the same

Explanation / Answer

12 ) P = I × V = R × I2 = V2 R

since voltage is same for both bulbs , so power is inversly proptional to square of resistance

p1/p2 = r2/r1 = 75/150 = r2/r1 = 1/2 = r2/r21

r1 = 2 * r2

so 75 w bulb has twice of the resistance of 150 w bulb

Ans B

13 ) We know r = rho * l /a , we see R esistance is proptional to length

r1/r2 = l1/l2 = r1/r2 = L/L/2 = R1/R2 = 2

R1 = 2 R2

R2 = R1/2

Ans d

14 .) In a series circuit containing any number of initial resistors connected to a
voltage source ... if a resistor is added in series ... then the total current
decreases, and the voltage drop across each initial resistor decreases.

In a series circuit the same current flows through each resistor. As the
total resistance increases (Rtotal) then the current (i) decreases, since
i = (Vs) / (Rtotal) , and Vs is the source voltage. A lower current means
a decrease in the voltage drop across each resistor: Vresistor = i R

Ans A

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