I\'m having trouble with this Physics problem. U se the circuit shown to answer

Question

I'm having trouble with this Physics problem.  Use the circuit shown to answer the following questions. At time t = 0 the switch in the circuit is closed. Answer the following questions after the switch in the circuit is closed.

b) What is the voltage drop across R6?

c) What is the current through R4?

d) Suppose instead of resistors, lightbulbs were used in the circuit. How would the brightness of R1 change if R6 were removed from the circuit?Suppose instead of resistors, lightbulbs were used in the circuit. How would the brightness of R1 change if R6 were removed from the circuit?

e) What is the new voltage drop across R1 after R6 is removed?

Please help me!  I really want to understand how to find these values.  Thank you so much for your help!

I'm having trouble with this Physics problem. Use the circuit shown to answer the following questions. At time t = 0 the switch in the circuit is closed. Answer the following questions after the switch in the circuit is closed. a) What is the equivalent resistance of the circuit given R1 = 7 Omega, R2 = 5 Omega, R3 = 4 Omega, R4 = 5 Omega, R5 = 7 Omega, and R6 = 2 Omega and a 7 V battery? - I calculated this all out and I got the answer to be 124/9 ohms, which is correct. This is where I'm having trouble b) What is the voltage drop across R6? c) What is the current through R4? d) Suppose instead of resistors, lightbulbs were used in the circuit. How would the brightness of R1 change if R6 were removed from the circuit?Suppose instead of resistors, lightbulbs were used in the circuit. How would the brightness of R1 change if R6 were removed from the circuit? e) What is the new voltage drop across R1 after R6 is removed?

Explanation / Answer

Current in circuit = 7/(124/9) = 0.508A

b.) Pot. Drop upto R1 and R2 = 7 - 0.508*(7+5) = 0.904 V

which is the potential diff. accross R6 = 0.904 A

c.) eq. Resistance of R3 R4 and R5 = 16 ohm, R6 = 2ohms

therefore current in R4 = 0.508*2/(16+2) = 0.0564 A

d.) The net resistance of circuit will increase => current through R1 will Decrease => Brightness will decrease.

e.)Net circuit resistance = 28 ohm(all in series)

voltage drop = 7*7/28 = 1.75 V

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