I got a circuit below and just wondering if i got the concept of open circuit correct. So i understand that the 10V and 2.7k ohm resistor has no drop and no current between them and the rest of the resistors will experience voltage drops from the 24V source correct? Its asking me to find the potential difference between point b and point e. So would i just find the potential at point b and just subtract it by 10? from what i learned that there is still voltage across the wire just no current so the potential at point e would be 10 volts? correct? and also what would a volt meter read if i put it between D and C? is it 10V - the drop from point C to F? or is it just 10 volts? Thanks for the help!
You are correct about no drop across the 2.7k because that part of the circuit is open. The other part of the circuit is the closed loop indicated as a-b-c-f-e and includes the 24v source and 4 resistors marked 3.9k, 1.2k, 800 ohms, and 100 ohms. There is current flowing through this loop. The total voltage drop around the loop is 24 volts and the drop across each resistor is proportional to it's value. The higher the resistor value, the higher the drop across it. So to figure the drop across each resistor use this calculation: Voltage Drop = (Resistor Value/Total Resistance) X 24 volts You want the drop between points b and e. The effective resistance from b to e is the sum of 1.2k + 800 ohms + 100 ohms = 2100 ohms. The total resistance of the loop is 6000 ohms, so the drop from b to e is (2100/6000) X 24 = 8.4 volts. The drop across the 3.9k is (3900/6000) X 24 = 15.6 volts. Notice that 8.4 + 15.6 = 24 volts, the total drop around the loop.
thanks for the help. Sorry the writing at the bottom left of the circuit is actually a G. So the question was asking point B with respect to E where the open circuit is. So would it just be point b potential is (24V) - (the drop from the 3.9k resistor which is 15.6 V) = 8.4V. So to find out the potential difference between B and E would it just be 8.4V(potential at B) - 10V (potential at E) = -1.6V ?
Just noticed that there are TWO points marked "e" in your diagram. I calculated Vbe using the "e" on the left side, near the 24 volt source. Is that what you wanted?