Hi, as seen in the problem below, I need to find the voltage drop across the resistors when the switches are open. I know that voltage across a break is equal to the supplied voltage. But how about across a resistor?

Thanks for any help.

 

E = IR. Since I = 0, so does E.

 

When the switches are open the resistors are not connected to anything so they don't get anything. Answer is 0

 

For this exercise the answer is 0V. Practically, the switch is not infinite impedance when open, so there will be a VERY tiny voltage drop.

 

Hi, as seen in the problem below, I need to find the voltage drop across the resistors when the switches are open. I know that voltage across a break is equal to the supplied voltage. But how about across a resistor?

The kind of break you are talking about is a completely different situation.
For instance, consider the break caused by the open S1. There is voltage
across it because the terminals of S1 are connected appropriately with a
voltage source.

However, both sides of the resistors are not in contact with the voltage
source when the switches are open, so no voltage can appear across them.

BTW, none of questions a, b, or c actually ask what happens when both
switches are open!

 

for the first case.
v1 = 30/3 v2=0
for case 2
v1=0; v2=20
for case 3 (apply superposition)
v1= 10/r
v2=20/r
i solved this by mere observation if any correction by any one please reply

 

all good mr recca, except i like V1 = 10v and V2 = 20v for case a.

 

yeah i see that now,
wonder whether it was negligence or a typo on my part.
thanks anyways.

 

there is no voltage across that resistor simply because it is not connected "PROPERLY" to any voltage source in this circuit.hence no voltage is "experienced" by that resistor. i=v/r,v=0.so i=0.

 

there are 2 loops........
consider 2 loop current flowing thru each.......

[B]case1[/B]

the 2nd loop is opened hence there is no current thru it .......
hence consider only the 1st loop.......with R and 2R resistors......and Vs1.......
see the resistance......
one is double of the other one.
hence vltge drop across 2R will be DOUBLE of that across R....

since V1 = IR...........(1)

and V2 = I*(2R)

HENCE.... V2/2= IR..........(2)
FROM (1) & (2)

V1=V2/2
2V1=V2
APPLY KVL....
VS1=V1+V2
30= V1+2V1
30=3V1
THEREFORE V1=10V

HENCE V2=20V...........SINCE V2=2*V1


CASE2

SIMILARLY.
FOR THE SECOND CASE....
1ST LOOP WILL NOT BE PRESENT......
AND R IS PARALLEL WITH VS2...
HENCE V2=VS2=10 OR 20 WATEVER IS GIVEN.....

 

Since the drop across 2R is 0, does it mean that if I measure the voltage across the open switch S2, I'll get all 20v?

 

Yes. The open switch is an insanely large resistance, so it drops all the voltage.

 

Thats a much better explanation than "the whole voltage appears at the terminals" for people who are confused.

 

there is no way to figure it if you do not know what the resistance or amps are. unless you figure R and 2r as anything you pick out. E/I*R so you have to know 2 of these