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Why doesn't using more appliances at home decrease the electricity bill?
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I know that at home the electric circuits are parallel, and this explains why if one appliance (eg bulb) fails, everything else continues to work, but if more devices are added in parallel to each other, their combined resistance should decrease, and thus the total power supplied by them should increase, as $mathrmPower = V^2/ R$. It doesn't look like that's the case, what am I getting wrong?
electricity electric-circuits electrical-resistance power
edited 1 hour ago
lr1985
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asked 1 hour ago
Rix Vii
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up vote
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I know that at home the electric circuits are parallel, and this explains why if one appliance (eg bulb) fails, everything else continues to work, but if more devices are added in parallel to each other, their combined resistance should decrease, and thus the total power supplied by them should increase, as $mathrmPower = V^2/ R$. It doesn't look like that's the case, what am I getting wrong?
electricity electric-circuits electrical-resistance power
edited 1 hour ago
lr1985
33419
asked 1 hour ago
Rix Vii
1067
Let us continue this discussion in chat.
– Harshit Joshi
51 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I know that at home the electric circuits are parallel, and this explains why if one appliance (eg bulb) fails, everything else continues to work, but if more devices are added in parallel to each other, their combined resistance should decrease, and thus the total power supplied by them should increase, as $mathrmPower = V^2/ R$. It doesn't look like that's the case, what am I getting wrong?
electricity electric-circuits electrical-resistance power
edited 1 hour ago
lr1985
33419
asked 1 hour ago
Rix Vii
1067
I know that at home the electric circuits are parallel, and this explains why if one appliance (eg bulb) fails, everything else continues to work, but if more devices are added in parallel to each other, their combined resistance should decrease, and thus the total power supplied by them should increase, as $mathrmPower = V^2/ R$. It doesn't look like that's the case, what am I getting wrong?
electricity electric-circuits electrical-resistance power
electricity electric-circuits electrical-resistance power
edited 1 hour ago
lr1985
33419
asked 1 hour ago
Rix Vii
1067
edited 1 hour ago
lr1985
33419
asked 1 hour ago
Rix Vii
1067
edited 1 hour ago
lr1985
33419
edited 1 hour ago
lr1985
33419
edited 1 hour ago
lr1985
33419
33419
asked 1 hour ago
Rix Vii
1067
asked 1 hour ago
Rix Vii
1067
asked 1 hour ago
Rix Vii
1067
1067
Let us continue this discussion in chat.
– Harshit Joshi
51 mins ago
add a comment |Â
Let us continue this discussion in chat.
– Harshit Joshi
51 mins ago
Let us continue this discussion in chat.
– Harshit Joshi
51 mins ago
Let us continue this discussion in chat.
– Harshit Joshi
51 mins ago
add a comment |Â
2 Answers
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oldest
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up vote
3
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accepted
The power will increase .See the formula given carefully (V^2)/R .For constant voltage power is inversly propotional to resistance.As we add one more resistance in parallel overall resistance decreases and hence power increases.Or you can also think inthe way if overall resistance decreases then current increases and hence P=VI increses
answered 31 mins ago
Sourabh
969
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Sourabh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
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down vote
To minimize your power consumption and save on electricity bills, you actually want to maximize your home's equivalent resistance, for the reason you mention. This is done when every appliance is turned off, because each turned-off appliance has (approximately) infinite resistance (up to some current leakage). Adding a new turned turned-off appliance doesn't really create a new current path or decrease the equivalent resistance. Every time you turn on an appliance, you are closing a new circuit and decreasing your home's equivalent resistance, therefore increasing your power consumption.
answered 17 mins ago
tparker
21.2k142113
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The power will increase .See the formula given carefully (V^2)/R .For constant voltage power is inversly propotional to resistance.As we add one more resistance in parallel overall resistance decreases and hence power increases.Or you can also think inthe way if overall resistance decreases then current increases and hence P=VI increses
answered 31 mins ago
Sourabh
969
New contributor
Sourabh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
3
down vote
accepted
The power will increase .See the formula given carefully (V^2)/R .For constant voltage power is inversly propotional to resistance.As we add one more resistance in parallel overall resistance decreases and hence power increases.Or you can also think inthe way if overall resistance decreases then current increases and hence P=VI increses
answered 31 mins ago
Sourabh
969
New contributor
Sourabh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The power will increase .See the formula given carefully (V^2)/R .For constant voltage power is inversly propotional to resistance.As we add one more resistance in parallel overall resistance decreases and hence power increases.Or you can also think inthe way if overall resistance decreases then current increases and hence P=VI increses
answered 31 mins ago
Sourabh
969
New contributor
Sourabh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The power will increase .See the formula given carefully (V^2)/R .For constant voltage power is inversly propotional to resistance.As we add one more resistance in parallel overall resistance decreases and hence power increases.Or you can also think inthe way if overall resistance decreases then current increases and hence P=VI increses
answered 31 mins ago
Sourabh
969
New contributor
Sourabh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 31 mins ago
Sourabh
969
New contributor
Sourabh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 31 mins ago
Sourabh
969
answered 31 mins ago
Sourabh
969
969
New contributor
Sourabh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sourabh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sourabh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
up vote
2
down vote
To minimize your power consumption and save on electricity bills, you actually want to maximize your home's equivalent resistance, for the reason you mention. This is done when every appliance is turned off, because each turned-off appliance has (approximately) infinite resistance (up to some current leakage). Adding a new turned turned-off appliance doesn't really create a new current path or decrease the equivalent resistance. Every time you turn on an appliance, you are closing a new circuit and decreasing your home's equivalent resistance, therefore increasing your power consumption.
answered 17 mins ago
tparker
21.2k142113
add a comment |Â
up vote
2
down vote
To minimize your power consumption and save on electricity bills, you actually want to maximize your home's equivalent resistance, for the reason you mention. This is done when every appliance is turned off, because each turned-off appliance has (approximately) infinite resistance (up to some current leakage). Adding a new turned turned-off appliance doesn't really create a new current path or decrease the equivalent resistance. Every time you turn on an appliance, you are closing a new circuit and decreasing your home's equivalent resistance, therefore increasing your power consumption.
answered 17 mins ago
tparker
21.2k142113
add a comment |Â
up vote
2
down vote
up vote
2
down vote
To minimize your power consumption and save on electricity bills, you actually want to maximize your home's equivalent resistance, for the reason you mention. This is done when every appliance is turned off, because each turned-off appliance has (approximately) infinite resistance (up to some current leakage). Adding a new turned turned-off appliance doesn't really create a new current path or decrease the equivalent resistance. Every time you turn on an appliance, you are closing a new circuit and decreasing your home's equivalent resistance, therefore increasing your power consumption.
answered 17 mins ago
tparker
21.2k142113
To minimize your power consumption and save on electricity bills, you actually want to maximize your home's equivalent resistance, for the reason you mention. This is done when every appliance is turned off, because each turned-off appliance has (approximately) infinite resistance (up to some current leakage). Adding a new turned turned-off appliance doesn't really create a new current path or decrease the equivalent resistance. Every time you turn on an appliance, you are closing a new circuit and decreasing your home's equivalent resistance, therefore increasing your power consumption.
answered 17 mins ago
tparker
21.2k142113
answered 17 mins ago
tparker
21.2k142113
answered 17 mins ago
tparker
21.2k142113
answered 17 mins ago
tparker
21.2k142113
21.2k142113
add a comment |Â
add a comment |Â
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Let us continue this discussion in chat.
– Harshit Joshi
51 mins ago