What Voltage Makes the 6 Ohm Resistor Absorb 1.5 J/s?

[terryds]

Homework Statement

A circuit is arranged like above.
To make the energy absorbed by the 6 ohm resistor every second is 1.5 J, voltage X should be ... Volt

Homework Equations

V = IR

The Attempt at a Solution

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I really have no idea.
I just know that it's kinda like wheatstone bridge and the current that goes to the 5 ohm resistor is 0 ampere. But, it has nothing to do with the question.
Please help me

 

[gneill]

What is the better-known unit for joules/second?
What formulas do you know for calculating this quantity when resistors are involved?

You'll need to write equations for and solve the circuit to find the current through the 6 Ω resistor in terms of the battery voltage.

 

[terryds]

What is the better-known unit for joules/second?
What formulas do you know for calculating this quantity when resistors are involved?

You'll need to write equations for and solve the circuit to find the current through the 6 Ω resistor in terms of the battery voltage.

I know it's called Power
P = I^2 R

But, how should I start? How to determine the current that goes in 6 ohm resistor ?

 

[gneill]

What have you been taught about KVL and KCL and writing circuit equations?

edit:
By the way, what was your logic in deciding that the current through the 5 Ohm resistor is zero? (It is a correct conclusion, but I'm curious as to how your arrived there).

 

[terryds]

What have you been taught about KVL and KCL and writing circuit equations?

Okay, I'll try..
P = I^2 R
1.5 = I^2 6
I = 0.5 A

So, the current that goes in 6 ohm resistor is 0.5 Ampere
The current goes in 5 ohm resistor is 0 ampere

I'll use mesh analysis.. Mesh 1 is the upper section, mesh 2 is lower section, mesh 3 is the right section

KVL on mesh 1

-2(0.5-0) -6(0.5) + x = 0
-1-3+x = 0
x = 4 Volt

Is it correct?

 

[gneill]

Okay, I'll try..
P = I^2 R
1.5 = I^2 6
I = 0.5 A

So, the current that goes in 6 ohm resistor is 0.5 Ampere
The current goes in 5 ohm resistor is 0 ampere

Okay, that's an excellent approach, working backwards from the desired result.

If the current through the 5 Ohm resistor is zero (why?) then where must this 1/2 Amp current flow?

I'll use mesh analysis.. Mesh 1 is the upper section, mesh 2 is lower section, mesh 3 is the right section

KVL on mesh 1

-2(0.5-0) -6(0.5) + x = 0
-1-3+x = 0
x = 4 Volt

Is it correct?

Yes. You only needed to consider that one loop because you already determined the required current and KVL must hold for that one loop. (And there is no mesh current in loop 3 to so loop 1 is effectively isolated (again, how did you know this?)).

To be more complete you should note that since the power dissipated in the resistor is given by I²R, the current can have two values: +1/2 and -1/2 amps. Squaring the current makes either one a positive quantity. That will give you two possible values for the battery voltage.

 

[terryds]

Okay, that's an excellent approach, working backwards from the desired result.

If the current through the 5 Ohm resistor is zero (why?) then where must this 1/2 Amp current flow?

Yes. You only needed to consider that one loop because you already determined the required current and KVL must hold for that one loop. (And there is no mesh current in loop 3 to so loop 1 is effectively isolated (again, how did you know this?)).

To be more complete you should note that since the power dissipated in the resistor is given by I²R, the current can have two values: +1/2 and -1/2 amps. Squaring the current makes either one a positive quantity. That will give you two possible values for the battery voltage.

The current through 5 ohm resistor is zero because the crossing product resistors are equal.
It's like the Wheatstone bridge but arranged differently.
The 1/2 Ampere will flow downward the resistor since the current through 5 ohm resistor is zero. (If the current is supposed to be clockwise in mesh 1)

I think -1/2 Ampere is not possible, unless the current is supposed to be counterclockwise

(I take the negative sign as upward the resistor, and positive sign as downward the resistor)

By KVL,

- x + 6(0.5) + 2(0.5) = 0
x = 4

which yields the same answer.

Is that right ?

 

[gneill]

The current through 5 ohm resistor is zero because the crossing product resistors are equal.
It's like the Wheatstone bridge but arranged differently.

Okay, I just wanted to make sure that you grasped the reason (and could recognize the scenario in similar problems in the future).

The 1/2 Ampere will flow downward the resistor since the current through 5 ohm resistor is zero. (If the current is supposed to be clockwise in mesh 1)

I think -1/2 Ampere is not possible, unless the current is supposed to be counterclockwise

There is no restriction on the current direction ... it is governed by the circuit potentials. If you reverse the battery potential then all current directions will reverse, too.

(I take the negative sign as upward the resistor, and positive sign as downward the resistor)

By KVL,

- x + 6(0.5) + 2(0.5) = 0
x = 4

which yields the same answer.

Is that right ?

Sure. And if the current direction were reversed?

 

[terryds]

Okay, I just wanted to make sure that you grasped the reason (and could recognize the scenario in similar problems in the future).

There is no restriction on the current direction ... it is governed by the circuit potentials. If you reverse the battery potential then all current directions will reverse, too.

Sure. And if the current direction were reversed?

You mean like this ?

- x + 6(-0.5) + 2(-0.5) = 0
- x -3 - 1 = 0
x = -4

So the potential can be 4 or -4 volt right ??
The negative and positive sign just tells that the current can goes either clockwise or counterclockwise, right?

 

[gneill]

You mean like this ?

- x + 6(-0.5) + 2(-0.5) = 0
- x -3 - 1 = 0
x = -4

So the potential can be 4 or -4 volt right ??
The negative and positive sign just tells that the current can goes either clockwise or counterclockwise, right?

Right.