Would anybody care to answer the questions?
Circuit 1
Circuit 2
Circuit 1
Circuit 2
Electrical Circuit Questions
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An ideal closed switch or un-blown fuse has zero resistance, thus zero volts for V2 and V3 in the first circuit.
Circuit is working correctly so we can assume these:
V1 Battery: charged = 12 volts. Of course a real battery is not exactly 12 volts.
V2 Fuse: not blown = zero volts.
V3 Switch: closed = zero volts
V4 Bulb: not burnt out = gets battery voltage minus the drop across the resistor, so 6 volts.
In circuit 2, the switch is open, so the entire battery voltage will appear across the switch (V3). Everything downstream of the switch is turned off, zero volts.
Circuit is working correctly so we can assume these:
V1 Battery: charged = 12 volts. Of course a real battery is not exactly 12 volts.
V2 Fuse: not blown = zero volts.
V3 Switch: closed = zero volts
V4 Bulb: not burnt out = gets battery voltage minus the drop across the resistor, so 6 volts.
In circuit 2, the switch is open, so the entire battery voltage will appear across the switch (V3). Everything downstream of the switch is turned off, zero volts.
Last edited:
Thank You mk378. I appreciate it very much.
Originally Posted By: mk378
An ideal closed switch or un-blown fuse has zero resistance, thus zero volts for V2 and V3 in the first circuit.
Circuit is working correctly so we can assume these:
V1 Battery: charged = 12 volts. Of course a real battery is not exactly 12 volts.
V2 Fuse: not blown = zero volts.
V3 Switch: closed = zero volts
V4 Bulb: not burnt out = gets battery voltage minus the drop across the resistor, so 6 volts.
In circuit 2, the switch is open, so the entire battery voltage will appear across the switch (V3). Everything downstream of the switch is turned off, zero volts.
Yes that looks like a good answer to me and well explained. Nice work.
I'll just add that the assumption is an ideal volt meter that offers infinite resistance, so it doesn't steal any voltage when taking a measurement.
An ideal closed switch or un-blown fuse has zero resistance, thus zero volts for V2 and V3 in the first circuit.
Circuit is working correctly so we can assume these:
V1 Battery: charged = 12 volts. Of course a real battery is not exactly 12 volts.
V2 Fuse: not blown = zero volts.
V3 Switch: closed = zero volts
V4 Bulb: not burnt out = gets battery voltage minus the drop across the resistor, so 6 volts.
In circuit 2, the switch is open, so the entire battery voltage will appear across the switch (V3). Everything downstream of the switch is turned off, zero volts.
Yes that looks like a good answer to me and well explained. Nice work.
I'll just add that the assumption is an ideal volt meter that offers infinite resistance, so it doesn't steal any voltage when taking a measurement.
JHZR2
Staff member
Originally Posted By: mk378
An ideal closed switch or un-blown fuse has zero resistance, thus zero volts for V2 and V3 in the first circuit.
Circuit is working correctly so we can assume these:
V1 Battery: charged = 12 volts. Of course a real battery is not exactly 12 volts.
V2 Fuse: not blown = zero volts.
V3 Switch: closed = zero volts
V4 Bulb: not burnt out = gets battery voltage minus the drop across the resistor, so 6 volts.
In circuit 2, the switch is open, so the entire battery voltage will appear across the switch (V3). Everything downstream of the switch is turned off, zero volts.
+1
An ideal closed switch or un-blown fuse has zero resistance, thus zero volts for V2 and V3 in the first circuit.
Circuit is working correctly so we can assume these:
V1 Battery: charged = 12 volts. Of course a real battery is not exactly 12 volts.
V2 Fuse: not blown = zero volts.
V3 Switch: closed = zero volts
V4 Bulb: not burnt out = gets battery voltage minus the drop across the resistor, so 6 volts.
In circuit 2, the switch is open, so the entire battery voltage will appear across the switch (V3). Everything downstream of the switch is turned off, zero volts.
+1
Originally Posted By: KJSmith
Originally Posted By: mk378
V4 Bulb: not burnt out = gets battery voltage minus the drop across the resistor, so 6 volts.
Circuit 1
Looks like V4 is reading to ground.
Bulb condition does not matter.
And picture two would be a dead short to ground when the switch is closed.
Originally Posted By: mk378
V4 Bulb: not burnt out = gets battery voltage minus the drop across the resistor, so 6 volts.
Circuit 1
Looks like V4 is reading to ground.
Bulb condition does not matter.
And picture two would be a dead short to ground when the switch is closed.
Originally Posted By: eljefino
Originally Posted By: SHOZ
And picture two would be a dead short to ground when the switch is closed.
nope 'cuz V4 is a sample point not a load.
Then they used the wrong symbol and should show the volt meter in there. Unless it was so described in the original paper.
Originally Posted By: SHOZ
And picture two would be a dead short to ground when the switch is closed.
nope 'cuz V4 is a sample point not a load.
Then they used the wrong symbol and should show the volt meter in there. Unless it was so described in the original paper.
Eljefino is correct. V4 is a voltmeter and is not a short to ground. Much the same as V1 across the battery that reads its voltage.
Originally Posted By: mk378
An ideal closed switch or un-blown fuse has zero resistance, thus zero volts for V2 and V3 in the first circuit.
Circuit is working correctly so we can assume these:
V1 Battery: charged = 12 volts. Of course a real battery is not exactly 12 volts.
V2 Fuse: not blown = zero volts.
V3 Switch: closed = zero volts
V4 Bulb: not burnt out = gets battery voltage minus the drop across the resistor, so 6 volts.
In circuit 2, the switch is open, so the entire battery voltage will appear across the switch (V3). Everything downstream of the switch is turned off, zero volts.
Agreed. (Ducked, are you one of the four Yorkshiremen?)
An ideal closed switch or un-blown fuse has zero resistance, thus zero volts for V2 and V3 in the first circuit.
Circuit is working correctly so we can assume these:
V1 Battery: charged = 12 volts. Of course a real battery is not exactly 12 volts.
V2 Fuse: not blown = zero volts.
V3 Switch: closed = zero volts
V4 Bulb: not burnt out = gets battery voltage minus the drop across the resistor, so 6 volts.
In circuit 2, the switch is open, so the entire battery voltage will appear across the switch (V3). Everything downstream of the switch is turned off, zero volts.
Agreed. (Ducked, are you one of the four Yorkshiremen?)
Originally Posted By: SHOZ
And picture two would be a dead short to ground when the switch is closed.
No.
Ideal Volt meter has infinite internal resistance/V. Otherwise it'll load down the CUT and give a false reading.
And picture two would be a dead short to ground when the switch is closed.
No.
Ideal Volt meter has infinite internal resistance/V. Otherwise it'll load down the CUT and give a false reading.
I tried running SPICE on this and it wouldn't converge.
;^)
;^)
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