Please show all work. Thanks! A 6 V battery supplies current to the circuit show

Question

Please show all work. Thanks!

A 6 V battery supplies current to the circuit shown in Figure P28.14. When the double-throw switch S is open, as shown in the figure, the current in the battery is 1.00 mA. When the switch is closed in position 1, the current in the battery is 1.10 mA. When the switch is closed in position 2, the current in the battery is 1.90 mA. Find the resistances R1, R2, and R3.

Explanation / Answer

When switch is open 6/(R1+R2+R3) = 1*10^-3 =>R1 + R2 + R3 = 6000 When switch in position 1 6/(R1+0.5R2+R3) = 1.1*10^-3 {R2 and R2 comes in parallel} R1+0.5R2+R3 = 5454.54 When switch in position 2 6/(R1 + R2 ) = 1.9*10^-3 {R3 gets shorted} R1 + R2 = 3157.89 On solving the three equations we get R1 = 2066.97 Ohms R2 = 1090.92 Ohms R3 = 2842.11 Ohms

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