I-------- 2 Ω -------------------- -----I
I I
I ======= +
I === - 5V
I I
I--------- 1 Ω ------------------------
I I
I ======= +
I === - 5V
I I
I--------- 1 Ω -----------------------
this is how the battery looks like.. (idid my best on my drawing :)
i have to find current and voltage for EACH RESISTORS (THERE ARE TOTAL OF 3 RESISTORS)
PLEASE SHOW THE STEPS.
THANKS FOR YOUR HELP FOR ADVANCE@@@
I I
I ======= +
I === - 5V
I I
I--------- 1 Ω ------------------------
I I
I ======= +
I === - 5V
I I
I--------- 1 Ω -----------------------
this is how the battery looks like.. (idid my best on my drawing :)
i have to find current and voltage for EACH RESISTORS (THERE ARE TOTAL OF 3 RESISTORS)
PLEASE SHOW THE STEPS.
THANKS FOR YOUR HELP FOR ADVANCE@@@
-
I'll try and explain the method, but it's fiddly without being able to point at a diagram.
You find the currents first. Then you can find each voltage using V=IR for each resistor in turn.
You have to initially guess each current's direction - if you get it wrong it doesn't matter, the value just turns out negative.
Label the current through the upper cell i1.
You can draw it going up or down, but for convenience I would draw an arrow going upwards next to the cell. This would be the current direction if only the top cell were present. The emf is then positive when you work round the upper loop in an anticlockwise direction (see later).
Label the current through the lower cell i2.
You can draw it going up or down. For the same reasoner just explained, I would draw an upwards arrow.
Label the current through the middle resistor i3. Guess the direction - draw it going left, say.
i1 goes through the 2Ω going left
i2 goes through the bottom 1Ω going right
Consider the junction on the middle of the left side.
Current going into junction = i1 + i3
Current leaving junction = i2
Using Kirchhoff's 1st Law, current entering = current leaving
i1+ i3 = i2 (equation 1)
Go anticlockwise around upper loop and apply Kirchhoff's 2nd Law.
emf = sum of voltages (voltage = current x resistance)
5 = 2i1 - 1i3 (equation 2) (the minus sign is because we are gong 'against' i3 when we go anticlockwise)
Go anticlockwise around lower loop and apply Kirchhoff's 2nd Law.
emf = sum of voltages
5 = 1i3 + 1i2 (equation 3)
Using equation 1, we eliminate i2 from equation 3
5 = 1i3 + 1(i1+i3)
5 = 2i3 + i1
i1 = 5 - 2i3 (equation 4)
Combining equations 2 and 4
5 = 2(5-2i3) - 1i3
= 10 -4i3 -i3
5i3 = 5
i3 = 1A
From equation 4
i1 = 5 - 2x1
= 3A
From equation 1
i2 = 3 + 1 = 4A
Voltage across top resistor = i1R = 3x2 = 6V
Voltage across middle resistor = i3R =1x1 = 1V
Voltage across lower resistor = i2R =4x1 = 4V
For a basic introduction to Kirchhoff's 2 laws, you might find the videos in the link useful.
You find the currents first. Then you can find each voltage using V=IR for each resistor in turn.
You have to initially guess each current's direction - if you get it wrong it doesn't matter, the value just turns out negative.
Label the current through the upper cell i1.
You can draw it going up or down, but for convenience I would draw an arrow going upwards next to the cell. This would be the current direction if only the top cell were present. The emf is then positive when you work round the upper loop in an anticlockwise direction (see later).
Label the current through the lower cell i2.
You can draw it going up or down. For the same reasoner just explained, I would draw an upwards arrow.
Label the current through the middle resistor i3. Guess the direction - draw it going left, say.
i1 goes through the 2Ω going left
i2 goes through the bottom 1Ω going right
Consider the junction on the middle of the left side.
Current going into junction = i1 + i3
Current leaving junction = i2
Using Kirchhoff's 1st Law, current entering = current leaving
i1+ i3 = i2 (equation 1)
Go anticlockwise around upper loop and apply Kirchhoff's 2nd Law.
emf = sum of voltages (voltage = current x resistance)
5 = 2i1 - 1i3 (equation 2) (the minus sign is because we are gong 'against' i3 when we go anticlockwise)
Go anticlockwise around lower loop and apply Kirchhoff's 2nd Law.
emf = sum of voltages
5 = 1i3 + 1i2 (equation 3)
Using equation 1, we eliminate i2 from equation 3
5 = 1i3 + 1(i1+i3)
5 = 2i3 + i1
i1 = 5 - 2i3 (equation 4)
Combining equations 2 and 4
5 = 2(5-2i3) - 1i3
= 10 -4i3 -i3
5i3 = 5
i3 = 1A
From equation 4
i1 = 5 - 2x1
= 3A
From equation 1
i2 = 3 + 1 = 4A
Voltage across top resistor = i1R = 3x2 = 6V
Voltage across middle resistor = i3R =1x1 = 1V
Voltage across lower resistor = i2R =4x1 = 4V
For a basic introduction to Kirchhoff's 2 laws, you might find the videos in the link useful.