I need help with this problem:
Picture: http://www.webassign.net/hecht/p18-38.gi…
Problem: Given that the ammeter in the figure (at right) reads 1.4 A, what is the emf of the ideal dc source as indicated by the voltmeter?
So I tried a very strange way of solving this.. I figured that if I = 1.4A in that spot it must be twice as much after the 6 ohm resistor (since the resistance is half) so it must be 2.8 there and then the current right after the 4 ohm resistor must equal the sum of those two (i.e. 1.4 + 2.8= 4.2 due to node rule) and that the current after the 2 ohm resistor must be twice as much so it must be 8.4 and then I multiplied that by the net resistance of 10 ohm and got 84 V.. and then I divided it by 2 to get 42 cause I figured that 84 must be the total current and you need it only for one side??
But I'm pretty sure that's the wrong way to do it. 42 V is the right answer but my method is wrong?
Picture: http://www.webassign.net/hecht/p18-38.gi…
Problem: Given that the ammeter in the figure (at right) reads 1.4 A, what is the emf of the ideal dc source as indicated by the voltmeter?
So I tried a very strange way of solving this.. I figured that if I = 1.4A in that spot it must be twice as much after the 6 ohm resistor (since the resistance is half) so it must be 2.8 there and then the current right after the 4 ohm resistor must equal the sum of those two (i.e. 1.4 + 2.8= 4.2 due to node rule) and that the current after the 2 ohm resistor must be twice as much so it must be 8.4 and then I multiplied that by the net resistance of 10 ohm and got 84 V.. and then I divided it by 2 to get 42 cause I figured that 84 must be the total current and you need it only for one side??
But I'm pretty sure that's the wrong way to do it. 42 V is the right answer but my method is wrong?
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The current through the 6Ω resistor is 2.8 Amps. The current in the 4Ω and 2Ω resistors is 4.2 Amps which is the sum of the 6.Ω resistor current (2.8 A) plus the current in the 12Ω resistor (1.4 A)
The voltage across the 6Ω resistor is 16.8 Volts (V = I*R = 2.8*6Ω). The voltage across the 4Ω resistor is 16.8 volts (= 4Ω * 4.2 Amps) and the voltage across the 2Ω resistor is 8.4 volts ( = 2Ω * 4.2 Amps)
V = 16.8 + 16.8 + 8.4 = 42 Volts which is the sum of the 4Ω, 6Ω, and 2Ω resistor voltage drops.
The voltage across the 6Ω resistor is 16.8 Volts (V = I*R = 2.8*6Ω). The voltage across the 4Ω resistor is 16.8 volts (= 4Ω * 4.2 Amps) and the voltage across the 2Ω resistor is 8.4 volts ( = 2Ω * 4.2 Amps)
V = 16.8 + 16.8 + 8.4 = 42 Volts which is the sum of the 4Ω, 6Ω, and 2Ω resistor voltage drops.