Three identical light bulbs are connected in series, then are disconnected and arranged in parallel. For each of the scenarios below indicate what changes (if any) take place
A. Total resistance of the circuit
B. Total current of the circuit
C. Power dissipated by the circuit
D. Voltage used by one of the light bulbs
E. Resistance of one of the light bulbs
3 answers
I will be happy to check your thinking.
A. series R=r+r+r parallel 1/r=1/r+1/r+1/r
B. series R=R1+R2+R3 Parallel 1/R= 1/r1+1/r2+1/r3
C. p=1xV=90 mA x 9V=0.81W
D. 3(1/3) Ax120 V= 120 W)
Are those right so far ?
B. series R=R1+R2+R3 Parallel 1/R= 1/r1+1/r2+1/r3
C. p=1xV=90 mA x 9V=0.81W
D. 3(1/3) Ax120 V= 120 W)
Are those right so far ?
A. you get the idea, but results can be simplified further
Since the bulbs are identical each with resistance r, then
in series, R=3r,
in parallel, 1/R=(1/r+1/r+1/r)=3/r, in other words, R=r/3.
B. Current = i, governed by V=iR.
For light bulbs in household circuit, V is constant, so i=V/R.
Retry using R calculated in A.
C. Power is Vi
Use results obtained in B, and multiply by V to express power in terms of V and R.
D. V=ir
In series, voltages is equally dropped over the three bulbs, so voltage drop of each is V/3.
In parallel, the voltage is the same across all bulbs, so V.
E. I'll leave that to you.
Since the bulbs are identical each with resistance r, then
in series, R=3r,
in parallel, 1/R=(1/r+1/r+1/r)=3/r, in other words, R=r/3.
B. Current = i, governed by V=iR.
For light bulbs in household circuit, V is constant, so i=V/R.
Retry using R calculated in A.
C. Power is Vi
Use results obtained in B, and multiply by V to express power in terms of V and R.
D. V=ir
In series, voltages is equally dropped over the three bulbs, so voltage drop of each is V/3.
In parallel, the voltage is the same across all bulbs, so V.
E. I'll leave that to you.
1 answer