What are the values of the currents a very short time (i.e. nearly immediately)

Question

What are the values of the currents a very short time (i.e. nearly immediately) after switch S is closed? (a) i1
A

(b) i2
A What are the values of the currents a very short time (i.e. nearly immediately) after switch S is closed? (a) i1
A

(b) i2
A (a) i1
A

(b) i2
A What are these values a long time later? (c) i1
A

(d) i2
A The switch is then reopened. Just then, what are the values of the currents? (e) i1
A

(f) i2
A What are these values a long time later? (g) i1
A

(h) i2
A What are these values a long time later? (c) i1
A

(d) i2
A The switch is then reopened. Just then, what are the values of the currents? (e) i1
A

(f) i2
A What are these values a long time later? (g) i1
A

(h) i2
A (c) i1
A

(d) i2
A (e) i1
A

(f) i2
A (g) i1
A

(h) i2
A

Explanation / Answer

a) When the switch is closed, the current through the resistors R1 and R2 are same resistors R1 and R2 are in series, so resultant is (R1+R2)          i1 = e / (R1 +R2)             =120 / (10+17)              = 4.44A b) I2 - 4.44A c) When the swithc is closed for loner time the equation of current is       e - i1 R1 -i2 R2 = 0    e - i1 R1 - (i1-i2)R3 = 0 solving the equation for current through R1     i1 = e (R2+R3) / [ R1R2 + R2 R3 + R3R1 ]         = (120)(10+17) / [ (10)(17)+(17)(30)+(30)(10)]         = 5.755A d) current through the resistor 2 is            i2 = eR3 / [ R1R2 + R2 R3 + R3R1 ]                 = (120)(30) / [ (10)(17)+(17)(30)+(30)(10)]                 = 3.673A e) When the switch is opened i1 = 0 as no current flow through this resistor. f) When the switch is open the current flow through the resistor r2 and is the diffenrce of part c and part d          = 5.755 - 3.673 = 2.08A g) For a long time the current i1 = 0 h) after long the time current decays and i2 = 0 .                 = 3.673A e) When the switch is opened i1 = 0 as no current flow through this resistor. f) When the switch is open the current flow through the resistor r2 and is the diffenrce of part c and part d          = 5.755 - 3.673 = 2.08A g) For a long time the current i1 = 0 h) after long the time current decays and i2 = 0 .        

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