We define the properties of parallel and complex circuits. We examine how to find the total voltage, current, and resistance of parallel and complex circuits. We also use a V.I.R. chart to find these things across individual resistors within a circuit.
equivalent resistance (Req) - the total resistance of a collection of resistors; for a series circuit, is equal to the sum of the resistances of the individual resistors. For a parallel circuit, the reciprocal of the equivalent resistance is equal to the reciprocal of each of the resistances of the individual resistors.
Resistors In Series And Parallel Practice Problems Pdf Download
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We learn about the properties of series circuits. We examine how to find the total voltage, current, and resistance of a series circuit and learn how to find these things across individual resistors within the circuit.
Here two transistors are in series. In figure (iii) total resistance will be less than individual resistances as they are connected parallel. Higher resistance produces more heat hence option c) is the right answer.
Thus, resistance of parallel combination is equal to resistance of resistors in series. So, potential difference across 20 resistance will be same as potential difference across the other two resistors which are connected in parallel.
29. Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit another set of three bulbs of the same wattage are connected in parallel to the same source.
(Resistance of the bulbs in series will be three times the resistance of single bulb. Hence, the current in the series combination will be one-third compared to current in each bulb in parallel combination. The parallel combination bulbs will glow more brightly.
A resistor network that is a combination of parallel and series connections can be broken up into smaller parts that are either one or the other. Some complex networks of resistors cannot be resolved in this manner, requiring more sophisticated circuit analysis. Generally, the Y-Δ transform, or matrix methods can be used to solve such problems.[4][5][6]
Practical resistors have a series inductance and a small parallel capacitance; these specifications can be important in high-frequency applications. In a low-noise amplifier or pre-amp, the noise characteristics of a resistor may be an issue.
Through-hole components typically have "leads" (pronounced /liːdz/) leaving the body "axially", that is, on a line parallel with the part's longest axis. Others have leads coming off their body "radially" instead. Other components may be SMT (surface mount technology), while high power resistors may have one of their leads designed into the heat sink.
Early resistors were made in more or less arbitrary round numbers; a series might have 100, 125, 150, 200, 300, etc.[30] Early power wirewound resistors, such as brown vitreous-enameled types, were made with a system of preferred values like some of those mentioned here. Resistors as manufactured are subject to a certain percentage tolerance, and it makes sense to manufacture values that correlate with the tolerance, so that the actual value of a resistor overlaps slightly with its neighbors. Wider spacing leaves gaps; narrower spacing increases manufacturing and inventory costs to provide resistors that are more or less interchangeable.
The former EIA-96 marking system now included in IEC 60062:2016 is a more compact marking system intended for physically small high-precision resistors. It uses a two-digit code plus a letter (a total of three alphanumeric characters) to indicate 1% resistance values to three significant digits.[31] The two digits (from "01" to "96") are a code that indicates one of the 96 "positions" in the standard E96 series of 1% resistor values. The uppercase letter is a code that indicates a power of ten multiplier. For example, the marking "01C" represents 10 kOhm; "10C" represents 12.4 kOhm; "96C" represents 97.6 kOhm.[32][33][34][35][36]
That's half the battle towards understanding the difference between series and parallel. We also need to understand how current flows through a circuit. Current flows from a high voltage to a lower voltage in a circuit. Some amount of current will flow through every path it can take to get to the point of lowest voltage (usually called ground). Using the above circuit as an example, here's how current would flow as it runs from the battery's positive terminal to the negative:
Notice that in some nodes (like between R1 and R2) the current is the same going in as at is coming out. At other nodes (specifically the three-way junction between R2, R3, and R4) the main (blue) current splits into two different ones. That's the key difference between series and parallel!
From the positive battery terminal, current flows to R1... and R2, and R3. The node that connects the battery to R1 is also connected to the other resistors. The other ends of these resistors are similarly tied together, and then tied back to the negative terminal of the battery. There are three distinct paths that current can take before returning to the battery, and the associated resistors are said to be in parallel.
However, this method is only good for two resistors in one calculation. We can combine more than 2 resistors with this method by taking the result of R1 R2 and calculating that value in parallel with a third resistor (again as product over sum), but the reciprocal method may be less work.
We need to be a little more careful when we combine resistors of dissimilar values in parallel where total equivalent resistance and power ratings are concerned. It should be completely obvious to the reader, but...
Adding capacitors in parallel is like adding resistors in series: the values just add up, no tricks. Why is this? Putting them in parallel effectively increases the size of the plates without increasing the distance between them. More area equals more capacitance. Simple.
In a nutshell they add just like resistors do, which is to say they add with a plus sign when in series, and with product-over-sum when in parallel. The tricky part comes when they are placed close together so as to have interacting magnetic fields, whether intentionally or not. For this reason, it is preferable to have a single component rather than two or more, though most inductors are shielded to prevent interacting magnetic fields.
Question 26.If a person has five resistors each of value \(\frac 1 5 \) Ω, then the maximum resistance he can obtain by connecting them is(a) 1 Ω(b) 5 Ω(c) 10 Ω(d) 25 Ω (2020)Answer:(a) The maximum resistance can be obtained from a group of resistors by connecting them in series. Thus,Rs = \(\frac15+\frac15+\frac15+\frac15+\frac15\) 1 Ω
Question 28.The maximum resistance which can be made using four resistors each of resistance \(\frac 1 2 \) Ω is(a) 2 Ω(b) 1 Ω(c) 2.5 Ω(d) 8 Ω (2020)Answer:(a) The maximum resistance can be produced from a group of resistors by connecting them in series.Thus, Rs = \(\frac 1 2 \) Ω + H \(\frac 1 2 \) Ω + \(\frac 1 2 \) Ω + \(\frac 1 2 \) Ω = 2 Ω
Question 29.Three resistors of 10 Ω, 15 Ω and 5 Ω are connected in parallel. Find their equivalent resistance. (Board Term I, 2014)Answer:Here, R1 = 10 Ω, R2 =15 Ω, R3 = 5 Ω.In parallel combination, equivalent resistance, (Req) is given by
Question 30.List the advantages of connecting electrical devices in parallel with an electrical source instead of connecting them is series. (Board Term I, 2013)Answer:(a) When a number of electrical devices are connected in parallel, each device gets the same potential difference as provided by the battery and it keeps on working even if other devices fail. This is not so in case the devices are connected in series because when one device fails, the circuit is broken and all devices stop working.
Question 31.Show how would you join three resistors, each of resistance 9 Ω so that the equivalent resistance of the combination is (i) 13.5 Ω, (ii) 6 Ω (2018)Answer:(i) The resistance of the series combination is higher than each of the resistances. A parallel combination of two 9 Ω resistors is equivalent to 4.5 Ω. We can obtain 13.5 Ω by coupling 4.5 Ω and 9 Ω in series. So, to obtain 13.5 Ω, the combination is as shown in figure (a). 2ff7e9595c
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