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# AS Level Physics question on electricity. Help needed! watch

1. A 12.0 V battery is connected with its positiveterminal to A and negative terminal to B.(i) Calculate the current in the 8.0 Ω resistor.......answer........... ....................... A

The answer is 1.5A and to get it you do 12v/8 ohms to get 1.5A (from mark scheme). But I'm very confused as to why my method and answer from my method is wrong.

My method is this:

1/8+1/4= 3/8.

3/8^-1 = 8/3 or 2.67

12v/2.67ohms = 4.5A

2:1 ratio therefore 4.5/3 is 1.5

1.5*2 is 3A.

Sorry if this seems like a stupid question but I'm just VERY confused...

Thanks!
2. (Original post by Smile Generator)

A 12.0 V battery is connected with its positiveterminal to A and negative terminal to B.(i) Calculate the current in the 8.0 Ω resistor.......answer........... ....................... A

The answer is 1.5A and to get it you do 12v/8 ohms to get 1.5A (from mark scheme). But I'm very confused as to why my method and answer from my method is wrong.

My method is this:

1/8+1/4= 3/8.

3/8^-1 = 8/3 or 2.67

12v/2.67ohms = 4.5A

2:1 ratio therefore 4.5/3 is 1.5

1.5*2 is 3A.

Sorry if this seems like a stupid question but I'm just VERY confused...m

Thanks!
Well as voltage is the same, between a parallel combination, it will be 12V across the 8 ohm (aswell as the diode & 4 ohm).

So therefore using I = V / R = 12 / 8 = 1.5A

As to why your method doesn't quite work, I'm not quite sure to be honest, that's the method I would've used if it was a more complex circuit (I wonder if the diode has something to do with it?), though the quick method I posted above does work.
3. (Original post by JohnnyDavidson)
Well as voltage is the same, between a parallel combination, it will be 12V across the 8 ohm (aswell as the diode & 4 ohm).

So therefore using I = V / R = 12 / 8 = 1.5A

As to why your method doesn't quite work, I'm not quite sure to be honest, that's the method I would've used if it was a more complex circuit (I wonder if the diode has something to do with it?), though the quick method I posted above does work.
Ahh, thanks so much! You're a life saver! ^_^
4. (Original post by Smile Generator)

A 12.0 V battery is connected with its positiveterminal to A and negative terminal to B.(i) Calculate the current in the 8.0 Ω resistor.......answer........... ....................... A

The answer is 1.5A and to get it you do 12v/8 ohms to get 1.5A (from mark scheme). But I'm very confused as to why my method and answer from my method is wrong.

My method is this:

1/8+1/4= 3/8.

3/8^-1 = 8/3 or 2.67

12v/2.67ohms = 4.5A

2:1 ratio therefore 4.5/3 is 1.5

1.5*2 is 3A.

Sorry if this seems like a stupid question but I'm just VERY confused...

Thanks!
The problem is with your ratio. 2:1 is the ratio of current through the 4Ω resistor to current through the 8Ω resistor, not the other way around.

In general, for two parallel resistors, I2/I1 = R1/R2. This can be derived from Ohm's law using the fact that the potential difference across each will be the same.
5. (Original post by Arbolus)
The problem is with your ratio. 2:1 is the ratio of current through the 4Ω resistor to current through the 8Ω resistor, not the other way around.

In general, for two parallel resistors, I2/I1 = R1/R2. This can be derived from Ohm's law using the fact that the potential difference across each will be the same.
Oh yeah... Thanks so much! I feel like a moron for not being able to figure that out before...
6. (Original post by Smile Generator)

A 12.0 V battery is connected with its positiveterminal to A and negative terminal to B.(i) Calculate the current in the 8.0 Ω resistor.......answer........... ....................... A

The answer is 1.5A and to get it you do 12v/8 ohms to get 1.5A (from mark scheme). But I'm very confused as to why my method and answer from my method is wrong.

My method is this:

1/8+1/4= 3/8.

3/8^-1 = 8/3 or 2.67

12v/2.67ohms = 4.5A

2:1 ratio therefore 4.5/3 is 1.5

1.5*2 is 3A.

Sorry if this seems like a stupid question but I'm just VERY confused...

Thanks!
What the last guy said... also you've made it more complicated than you needed to.
The PD across the 8ohm resistor is 12V so you just need to do
I=V/R

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