The "earth" symbol on the bottom line tells you that the potential (voltage) is zero volts at all points on it.
V1 is the potential at that point compared with zero.
So it could be the pd across the 8 ohm resistor.
V1 is the potential at that point compared with zero.
So it could be the pd across the 8 ohm resistor.
Original post by Stonebridge
The "earth" symbol on the bottom line tells you that the potential (voltage) is zero volts at all points on it.
V1 is the potential at that point compared with zero.
So it could be the pd across the 8 ohm resistor.
V1 is the potential at that point compared with zero.
So it could be the pd across the 8 ohm resistor.
I understand V1 represents the voltage of the 8 ohm resistor. Maybe I should have phrased the question differently:
If we were to measure the voltage at the red dot, would the value we obtain refer to the voltage AT that dot or would it be referring to the voltage across the 2ohm resistor next to it? Voltage = potential difference no?
thanks
Original post by sabre2th1
I understand V1 represents the voltage of the 8 ohm resistor. Maybe I should have phrased the question differently:
If we were to measure the voltage at the red dot, would the value we obtain refer to the voltage AT that dot or would it be referring to the voltage across the 2ohm resistor next to it? Voltage = potential difference no?
thanks
If we were to measure the voltage at the red dot, would the value we obtain refer to the voltage AT that dot or would it be referring to the voltage across the 2ohm resistor next to it? Voltage = potential difference no?
thanks
The absolute "voltage" (at a point) in a circuit is meaningless unless it is measured wrt to some base value, usually taken as zero on the earthed line.
When you measure "the voltage at the red dot" you are measuring the potential difference between earth and that point. So clearly that isn't the pd across the 2 ohm resistor.
So solve the circuit using the network rules and the pd across the 8 ohm resistor (from this analysis) will give you the value of V1
Original post by sabre2th1
I understand V1 represents the voltage of the 8 ohm resistor. Maybe I should have phrased the question differently:
If we were to measure the voltage at the red dot, would the value we obtain refer to the voltage AT that dot or would it be referring to the voltage across the 2ohm resistor next to it? Voltage = potential difference no?
thanks
If we were to measure the voltage at the red dot, would the value we obtain refer to the voltage AT that dot or would it be referring to the voltage across the 2ohm resistor next to it? Voltage = potential difference no?
thanks
If the circuit is referenced to ground potential (the earth symbol) it's a big hint that all voltages are wrt to ground.
When you construct your mesh equations, the loop currents (standard convention) will flow from a +ve potential to ground or from ground to a lower -ve potential.
Original post by Stonebridge
The absolute "voltage" (at a point) in a circuit is meaningless unless it is measured wrt to some base value, usually taken as zero on the earthed line.
When you measure "the voltage at the red dot" you are measuring the potential difference between earth and that point. So clearly that isn't the pd across the 2 ohm resistor.
So solve the circuit using the network rules and the pd across the 8 ohm resistor (from this analysis) will give you the value of V1
When you measure "the voltage at the red dot" you are measuring the potential difference between earth and that point. So clearly that isn't the pd across the 2 ohm resistor.
So solve the circuit using the network rules and the pd across the 8 ohm resistor (from this analysis) will give you the value of V1
In the below circuit (its slightly different from the one I posted earlier), if V1 = 28V, am I correct in saying the voltage across the 2 ohm resistor is 12 volts?
Because the voltage before the 2 ohm resistor is 40, and 40-12= 28 V which is the voltage, V1 after the 2 ohm resistor..
Original post by sabre2th1
In the below circuit (its slightly different from the one I posted earlier), if V1 = 28V, am I correct in saying the voltage across the 2 ohm resistor is 12 volts?
Because the voltage before the 2 ohm resistor is 40, and 40-12= 28 V which is the voltage, V1 after the 2 ohm resistor..
Because the voltage before the 2 ohm resistor is 40, and 40-12= 28 V which is the voltage, V1 after the 2 ohm resistor..
Yes. The voltage (V2, say) at a point in the top left corner of the circuit must be +40V
If V1 is 28V then the pd across the 2 ohm is 40V - 28V
The pds across the 2ohm and the 8ohm must add to give 40V in that left hand loop.
And it's the same idea on the right hand side of the circuit.
Original post by sabre2th1
In the below circuit (its slightly different from the one I posted earlier), if V1 = 28V, am I correct in saying the voltage across the 2 ohm resistor is 12 volts?
Because the voltage before the 2 ohm resistor is 40, and 40-12= 28 V which is the voltage, V1 after the 2 ohm resistor..
Because the voltage before the 2 ohm resistor is 40, and 40-12= 28 V which is the voltage, V1 after the 2 ohm resistor..
Yes.
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