Two 110-V light bulbs, one \"25 W\" and the other 100 W\" are connected in serie

Question

Two 110-V light bulbs, one "25 W" and the other 100 W" are connected in series to a 110 V source. Then: the current in the 100-W bulb is greater than that in the 25-W bulb the current in the 100-W bulb is less than that in the 25 W bulb both bulbs will light with equal brightness each bulb will have a potential difference of 55 V none of the above A resistor is made out of a wire having a length L. When the ends of the wire are attached across the terminals of an ideal battery having a constant voltage V0 across its terminals, a current I flows through the wire. If the wire were cut in half, making two wires of length L/2, and both wires were attached across the terminals of the battery (the right ends of both wires attached to one terminal, and the left ends attached to the other terminal), how much current would the battery put out? 4I 2I I I/2 I/4 A 7.0-Ohm resistor is connected across the terminals of a battery having an internal emf of 10 V. If 0.50-A current flows, what is the internal resistance of the battery? 13 Ohm 20 Ohm 10.8 Ohm 27.0 Ohm

Explanation / Answer

1.solution

E. none of the above.

100w = V^2/R or I^2R
25w = V^2/R or I^2R

R(100w bulb) = V^2/100 = 110 x 110 / 100 = 121 ohms
R (25v bulb) = V^2/100 = 110 x 110 / 25 = 484 ohms
R (both bulbs) = 484 + 121 = 605 ohms

The current through both bulbs = 110 / 605 = 0.1818 amps

using watta = I^2R
(100w bulb) 0.1818 x 0.1818 x 121 = 3.9999 = 4w = lower than 100w
(25w bulb) 0.1818 x 0.1818 x 484 = 15.9968 = 16w = lower than 25w

2.solution

1st condition:-
=>Let the resistance of the wire of length L is R ,
=>By V = IR
=>I = V/R = Vo/R --------------(i)
2nd condition:-
=>By R = L/A
=>The new resistance of each length i.e. L/2 will be equal to R/2 , and the resistors are now connected in parallel:-
=>By 1/R(net) = 1/R1 + 1/R2
=>1/R(net) = 2/R + 2/R
=>R(net) = R/4
Thus by V = iR
=>Vo = i2 x R/4
=>i2 = 4 x [Vo/R]
=>i2 = 4I
=>(a)

3.solution

A) 13 ohms

The resistor is seeing a voltage of E = (IR) = 0.5 x 7 = 3.5V. across itself.
Thus (10 - 3.5) = 6.5V. is present across the battery terminals, when the circuit is connected.
R = (E/I) = 6.5/0.5 = 13 ohms apparent internal resistance

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