Isaac Physics question C4a.4 essential pre uni physics. Please help.
Both R 1 and R 2 are of resistance R.
The circuit shown consists of a resistor of resistance R in series with a parallel a parallel segment. There are 2 resistors of resistance R in series which is in turn parallel with a resistor of resistance R_2. this two connect at a node which then branch off into 2 branches in parallel with each other. One branch is a resistor of resistance R and the other is 2 resistors of resistance R in series with each other. There is a resistor of resistance R in parallel with these sections. This all connects in series to a resistor of resistor R.
Figure 1: A circuit is shown above. Assume the cell and wires are ideal.
Find the equivalent resistance in the circuit in terms of R.
(18*R)/(7)
Correct!
Find the potential across R 1 in terms of V.
The circuit shown consists of a resistor of resistance R in series with a parallel a parallel segment. There are 2 resistors of resistance R in series which is in turn parallel with a resistor of resistance R_2. this two connect at a node which then branch off into 2 branches in parallel with each other. One branch is a resistor of resistance R and the other is 2 resistors of resistance R in series with each other. There is a resistor of resistance R in parallel with these sections. This all connects in series to a resistor of resistor R.
Figure 1: A circuit is shown above. Assume the cell and wires are ideal.
Find the equivalent resistance in the circuit in terms of R.
(18*R)/(7)
Correct!
Find the potential across R 1 in terms of V.
Reply 1
Hopefully when calculating the overall resistance, you simplified the circuit to 4R/3 and R1 in parallel with 2R in series with V. This means that the voltage will be the same across R1 and 4/3R.
The parallel combination can be simplified further to 4R/7, so the voltage across the 4R/7 is the same as the voltage across R1. The voltage fraction is equivalent to V *(4/7) / (4/7 + 2R) = 2V/9.
The parallel combination can be simplified further to 4R/7, so the voltage across the 4R/7 is the same as the voltage across R1. The voltage fraction is equivalent to V *(4/7) / (4/7 + 2R) = 2V/9.
Reply 2
7
Original post by Alex_H_
Hopefully when calculating the overall resistance, you simplified the circuit to 4R/3 and R1 in parallel with 2R in series with V. This means that the voltage will be the same across R1 and 4/3R.
The parallel combination can be simplified further to 4R/7, so the voltage across the 4R/7 is the same as the voltage across R1. The voltage fraction is equivalent to V *(4/7) / (4/7 + 2R) = 2V/9.
The parallel combination can be simplified further to 4R/7, so the voltage across the 4R/7 is the same as the voltage across R1. The voltage fraction is equivalent to V *(4/7) / (4/7 + 2R) = 2V/9.
amazing!
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