A flashlight contains a battery of two cells in series, with a bulb of resistance 12 Ohms. The internal resistance of each cell
is 0.26 ohms. If the potential difference across the bulb is 2.88V, what is the emf of EACH cell?
1 answer:
Answer:
1.5024
Explanation:
Draw a diagram. Put the two cells in series. Now draw 3 resistors. Two of them equal 0.26 ohms each. The third one is the lightbulb which is 12 ohms.
R = 0.26 + 0.26 + 12 = 12.52
The bulb has a voltage of 2.88 volts across it. You can get the current from that.
i = E / R
i = 2.88 / 12 =
i = 0.24 amps.
Now you can get the voltage drop across the two cells.
E = ?
R = 0.26
i = 0.24 amps
E = 0.26 * 0.24
E = 0. 0624
Finally divide the 2.88 by 2 to get 1.44
Each cell has an emf of 1.44 + 0.0624 = 1.5024
You might be interested in
Answer:
mel mel 4 ever
Explanation:
Factor out 8 and then facotr and u get
8/9(9x+1)(9x-1
Answer:
49N
Explanation:
F=ma
We know the mass is 5kg, and since the ball is suspended on one cable, the acceleration is g, 9.8m/s^2
F=5kg*9.8m/s^2
= 49N
Hope this helps!
Answer:
I Will say the Answer is A
Explanation: