second question please
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B is 10 Ohms, per Kirchoff's Law, we have two equations with three unknowns (I1, I2, and B):
5 volts = 5 Ohms* I1 + 15 Ohms * ( I1 - I2)0 volts = (-I1 + I2) * 15 Ohms + I2* B ohms
Given that I2 = ⅓ Amp, we now have two equations with two unknowns (I1 and B).
Substituting and solving: B = 15/2 = 7.5 ohms.
Given that resistor B is 15/2, or 7.5, Ohm's, means that the parallel combination of the 15 ohm resistor and the 7.5 ohm resistor is the equivalent to a 5 ohm resistor, which means that we have, essentially two 5 ohm resistors in series, or the equivalent of one 10 ohm resistor. Thus the 5 volt battery "sees" a total resistance of 10 Ohm's. The 5 volt battery thus puts out a current of I = V/R = 5/10 = 0.5 amps.
Therefore the current flowing through the 5 ohm resistor is 0.5 amps, resulting in a 2.5 volt voltage drop. The current flowing through the parallel combination of the 15 ohm resistor and the 7.5 ohm resistor (aka resistor B) is also 0.5 amps, with ⅓ amps flowing through the 7.5 ohm resistor and ½ - ⅓ = 1/6 amps flowing through the 15 ohm resistor.
Finally, the 15 ohm resistor drops 2.5 volts (15 * 1/6) as does the 7.5 ohm resistor.
5 volts = 5 Ohms* I1 + 15 Ohms * ( I1 - I2)0 volts = (-I1 + I2) * 15 Ohms + I2* B ohms
Given that I2 = ⅓ Amp, we now have two equations with two unknowns (I1 and B).
Substituting and solving: B = 15/2 = 7.5 ohms.
Given that resistor B is 15/2, or 7.5, Ohm's, means that the parallel combination of the 15 ohm resistor and the 7.5 ohm resistor is the equivalent to a 5 ohm resistor, which means that we have, essentially two 5 ohm resistors in series, or the equivalent of one 10 ohm resistor. Thus the 5 volt battery "sees" a total resistance of 10 Ohm's. The 5 volt battery thus puts out a current of I = V/R = 5/10 = 0.5 amps.
Therefore the current flowing through the 5 ohm resistor is 0.5 amps, resulting in a 2.5 volt voltage drop. The current flowing through the parallel combination of the 15 ohm resistor and the 7.5 ohm resistor (aka resistor B) is also 0.5 amps, with ⅓ amps flowing through the 7.5 ohm resistor and ½ - ⅓ = 1/6 amps flowing through the 15 ohm resistor.
Finally, the 15 ohm resistor drops 2.5 volts (15 * 1/6) as does the 7.5 ohm resistor.
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