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Assistance is requested with the following electric circuit questions (see attachment).

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AS EQUATIONS and FIGURES ARE involved the PROPER SOLUTION IS PROVIDED IN ATTACHED docx & pdf FILE. Please find the attachment.


We will solve this circuit problem by using Kirchhoff's loop rule for this closed circuit loop. This rule says that sum of potential differences in any closed loop, involving emfs and those of resistances must equal to zero.

So according to this rule in above circuit sum of potentials must be zero i.e. sum of potentials due to emf source 48 V, emf source 36 V and potential across resistor of 12 ohm will be equal to zero.

But here, you will need to take care of direction and + & - signs associated with all three potentials. According to sign conventions for the Loop Rule, for the emf source sign of emf is '+' (i.e. +V) if you travel across emf source from '-' end to '+' end; otherwise for travel direction of '+' to '-' the emf will be negative(-).
In case of resistor if you travel in direction of current then the sign of value of resistance will be '-' (i.e. - ohm) ; Otherwise for direction opposite to current the sign of resistance will be '+'.
In given circuit the direction of current will be anti-clock wise direction as bigger source of emf is 48 V and convention current direction is '+' end to towards '-' end of emf source.( So the direction of Ix in circuit diagram of Question is wrong it should be opposite to what is drawn, correct direction is drawn here).
In above circuit let us start calculating potentials from top right corner (i.e. from top of resistor) and suppose that ...

Solution Summary

The solution assists with answering the given electric circuit questions.