In the circuit shown in the figure, the 6.0 ohm resistor is consuming energy at
ID: 2138345 • Letter: I
Question
In the circuit shown in the figure, the 6.0 ohm resistor is consuming energy at a rate of 23.0 J/s when the current through it flows as shown. A) Find the current through the ammeter in amps. B) What are the polarity and emf of the battery (E), assuming it has negligible internal resistance? In the circuit shown in the figure, the 6.0 ohm resistor is consuming energy at a rate of 23.0 J/s when the current through it flows as shown. A) Find the current through the ammeter in amps. B) What are the polarity and emf of the battery (E), assuming it has negligible internal resistance?Explanation / Answer
let's start by simplifying the circuit:
- the two ohm resistors combine in parallel to be replaced by one 10 ohm resistor
- the 10ohm, 19 ohm and 1 ohm resistors in the ammeter leg add to one 30 ohm resistor
- the 6 ohm and the 3 ohm resistors in the 25 V leg add up to a 9 ohm resistor
- the 17 ohm and 13 ohm resistors in the E leg add to one 30 ohm resistor
- since the 6 ohm resistor in the 25 V leg dissipates 23.0 watts, then since P = I^2 R,
the current through this leg is 2 amps
- since the current through the 25 V leg is 2 amps, then the voltage across the 25 V leg must be
25 - 2(9) = 7 V
1.) - this means the voltage across the ammeter leg is 7 V, which is also 7 V = (30 ohm)(Ia),
so the current through the ammeter leg is 7 / 30 amps from bottom to top
2.) From Kirchoffs Current law looking at the junction above the 25 V source, the current must
be flowing into the E leg from above because (Ie) = 2 + 7 / 30 = 67 / 30 where Ie = E leg current. So Ie = 67 / 30 amps where the ploarity is opposite of what the diagram shows with the minus on top and the + on the bottom. Since Ie = 67 / 30 amps and the voltage across the E leg has to be 7 V, then
E - (Ie)R = 7
E - (67 / 30)(30) = 7
E = 74 V