Question

1. Three identical resistors are connected in parallel to a single battery. The current through the resistors must be...

a. one-third of the current through the battery.
b. the same for the resistors as for the battery.
c. different for each resistor.
d. higher for one resistor and lower for the other two.

2. A 6Ω and a 12Ω resistor are connected in series with a 36 volt battery,. How many Joules per second are dissipated by the 12Ω resistor.

a. 14 watts
b. 24 watts
c. 48 watts
d. 64 watts

3. Six identical resistors are connected in series with a battery. The number of joules per second supplied by the battery is then determined. If a seventh resistor is added (in series) the number of joules per second supplied by the battery:

a. increases
b. decreases
c. stays the same
d. changes, but may increase or decrease

Answer 1

1.

When resistors are connected in parallel, then voltage in each of resistor will be equal and since all resistors are identical, So current in each of them will be equal and it will be one-third of the total current, Since if resistnace of one resistor is R then equivalent resistance is Req = R/3, So total current will be

I = V/(R/3) = 3V/R

Current in each resistor will be

I1 = I2 = I3 = V/R

Correct option is A.

2.

Power = Current^2*Resistance

We know that in series connection, Req = 6 + 12 = 18 Ohm

Ieq = V/Req = 36/18 = 2 Amp

Since series connection, So current in both resistor will be equal to total cureent, So power in 12 ohm resistor will be

P = i^2*R = 2^2*12 = 48 W

Correct option is C.

3.

Power = V*I

I = V/Req

Initially

Req = 6R

then , P = V*6R = 6VR

After 7th added

Req = 7R

Now, P = V*7R = 7VR

So, we can see that

power supplied will increase after addition of the 7th resistor.

Correct option is A.

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