# How To Solve Parallel Circuit Problems

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For Figure, the sum of the potential drop of each resistor and the voltage supplied by the voltage source should equal zero:$V - V_1 - V_2 - V_3 = 0,$$V = V_1 V_2 V_3,$$= IR_1 IR_2 IR_3,$$I = \frac = \frac.$Since the current through each component is the same, the equality can be simplified to an equivalent resistance, which is just the sum of the resistances of the individual resistors.

Any number of resistors can be connected in series.

The equivalent resistance of a set of resistors in a series connection is equal to the algebraic sum of the individual resistances.

In Figure, the current coming from the voltage source flows through each resistor, so the current through each resistor is the same.

The sum of the individual currents equals the current that flows into the parallel connections.

Resistors are said to be in series whenever the current flows through the resistors sequentially.The current through the circuit depends on the voltage supplied by the voltage source and the resistance of the resistors.For each resistor, a potential drop occurs that is equal to the loss of electric potential energy as a current travels through each resistor.According to Ohm’s law, the potential drop is the resistance in ohms $$(\Omega)$$.Since energy is conserved, and the voltage is equal to the potential energy per charge, the sum of the voltage applied to the circuit by the source and the potential drops across the individual resistors around a loop should be equal to zero:$\sum_^N V_i = 0.$This equation is often referred to as Kirchhoff’s loop law, which we will look at in more detail later in this chapter.They solve for total resistance and current, the current through each resistor, the voltage across each resistor, and the power dissipated.Z_T =((70 60j)(40-25j))/(110 35j) =((92.20/_40.60^text(o))(47.17/_-32.01^text(o)))/(115.4/_17.65^text(o)) (We do the product on the top first.) =((92.20xx47.17)/_(40.60^text(o)-32.01^text(o)))/(115.4/_17.65^text(o)) =(4349.074/_8.59^text(o))/(115.4/_17.65^text(o)) (Now we do the division.) =(4349.074)/115.4/_(8.59^text(o)-17.65^text(o)) =37.69/_-9.06^text(o) (We convert back to rectangular form.) =37.22-5.93j A 100\ Ω resistor, a 0.0200\ "H" inductor and a 1.20\ mu"F" capacitor are connected in parallel with a circuit made up of a 110\ Ω resistor in series with a 2.40\ mu"F" capacitor.The equivalent resistance of a combination of resistors depends on both their individual values and how they are connected.The simplest combinations of resistors are series and parallel connections (Figure).In a series circuit, the equivalent resistance is the algebraic sum of the resistances.The current through the circuit can be found from Ohm’s law and is equal to the voltage divided by the equivalent resistance.

## Comments How To Solve Parallel Circuit Problems

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