(a) Can the circuit shown above be reduced to a single resistorconnected to the

Question

(a) Can the circuit shown above be reduced to a single resistorconnected to the batteries? Explain. (Use the following variablesas necessary: V = 9, R1 = 2.8,and R2 = 0.5.)


(b) Calculate each of the unknowncurrents I1, I2,and I3 for the circuit. I1 A I2 A I3 A I1 A I2 A I3 A Use the figure below for parts (a) and (b). (a) Can the circuit shown above be reduced to a single resistorconnected to the batteries? Explain. (Use the following variablesas necessary: V = 9, R1 = 2.8,and R2 = 0.5.) (b) Calculate each of the unknowncurrents I1, I2,and I3 for the circuit. I1 A I2 A I3 A

Explanation / Answer

Given : .     You can only reduce a circuit to a single,equivalent resistance if there is only one battery in the circuit.If there are two batteries (or more) you must use Kirchhoffs looprules, which are a pain but there is no way around it. b )    from the figure using kirchoff's law    24 V = R1I1 + 4I1 + ( I1 - I2 ) 3    24 V = R1 I1 + 4 I1 + 3 I1 - 3I2 7.8 I1 - 3 I2 = 24 V ------ ( 1)    9 V = R2I2 + 5I2 + ( I2 - I1 ) 3    9 V = R2 I2 + 5 I2 + 3I2 - 3I1     8.5 I2 - 3 I1 = 9 V---------- ( 2 ) Solve for I 1 and I2    I3 = I1 - I2 =--- A Solve it .     You can only reduce a circuit to a single,equivalent resistance if there is only one battery in the circuit.If there are two batteries (or more) you must use Kirchhoffs looprules, which are a pain but there is no way around it. b )    from the figure using kirchoff's law    24 V = R1I1 + 4I1 + ( I1 - I2 ) 3    24 V = R1 I1 + 4 I1 + 3 I1 - 3I2 7.8 I1 - 3 I2 = 24 V ------ ( 1)    9 V = R2I2 + 5I2 + ( I2 - I1 ) 3    9 V = R2 I2 + 5 I2 + 3I2 - 3I1     8.5 I2 - 3 I1 = 9 V---------- ( 2 ) Solve for I 1 and I2    I3 = I1 - I2 =--- A Solve it

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