I would like to know why the voltage source is not used in the equation as given in this image below.
In the image the equation is:
\$-I1∙8Ω+I2∙4Ω=0\$
Why can't the equation be \$10-I1∙8Ω+I2∙4Ω=0\$ using KVL?
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2\$\begingroup\$ Voltage source ?? I do not see a voltage source in this schematic. I do see a current source though. \$\endgroup\$– BimpelrekkieCommented Apr 6, 2016 at 10:20
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\$\begingroup\$ @FakeMoustache:Oh that's right.That's why this doubt came.Now I can get the answer of user3219492.Should I delete this question? \$\endgroup\$– justinCommented Apr 6, 2016 at 10:22
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1\$\begingroup\$ @justin No, this is an acceptable question in that it shows an electronics problem, an attempted solution, and then a good corrected solution. Keeping it on the site would be worthwhile in my opinion. \$\endgroup\$– nanofaradCommented Apr 6, 2016 at 10:41
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2 Answers
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It can't be 10−I1∙8Ω+I2∙4Ω=0 because, in this equation, you are trying to add voltage (I1∙8Ω+I2∙4Ω) and current (10A). Just like how you can't add 1m and 1 sec. They are in different units. You are not allowed to do that.
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Because your source is a current source, which means that it will provide as much voltage as needed until current is 10A. Therefore source voltage is not a provided variable in this circuit.
As Justin puts it in its comment, a current source uses a mechanism similar to an ammeter to watch the output current. If output current is below required value, then it increases the output voltage. If output current is above required value, then it decreases the voltage. In other words, voltage is regulated so current can be constant.
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1\$\begingroup\$ :Do you mean to say that it uses an ammeter to check whether it has reached 10A? \$\endgroup\$– justinCommented Apr 6, 2016 at 10:34
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1\$\begingroup\$ Nice way of putting it. Not exactly an ammeter, but a current source contains a mechanism to measure current, and it increases the output voltage until it has reached 10A \$\endgroup\$– jmgonetCommented Apr 6, 2016 at 11:24