
Two resistances are $\mathrm{R}_{1}=100 \pm 3$ and $\mathrm{R}_{2}=200 \pm 4$ Ohms.
Calculate ${{\text{R}}_{\text{eq}}}$
(I) series combination
(ii)Parallel combination (In error limits)
Answer
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Hint: The potential drop in parallel through each resistor is the same. Each of the parallel resistors does not get the total current; they split it. The current entered by a parallel resistor combination is equal to the sum of the current in parallel through each resistor.
Formula used:
For series - $R={{R}_{1}}+{{R}_{2}}$
For Parallel - $R^{\prime}=\dfrac{R_{1} R_{1}}{R_{1}+R_{2}}$
Complete step by step solution:
(i) In a series resistor network, the individual resistors combine to give the series combination an equivalent resistance (RT). Without affecting the total resistance, current, or power of each resistor or circuit, the resistors in a series circuit can be interchanged. Resistance is directly proportional to the potential difference across a circuit's terminals. The potential difference is greater in sequences, thus offering more resistance.
The equivalent resistance of series
combination $R={{R}_{1}}+{{R}_{2}}=(100\pm 3)\text{ohm}+(200\pm 4)$ ohm
$=300 \pm 7$ ohm.
(ii) When two or more resistors are connected to each terminal of the other resistor or resistor so that both of their terminals are connected respectively, they are said to be connected together in parallel.
The equivalent resistance of parallel Combination
$R^{\prime}=\dfrac{R_{1} R_{1}}{R_{1}+R_{2}}=\dfrac{200}{3}=66.7$ ohm
Then, from $\dfrac{1}{R^{\prime}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}$
we get. $\dfrac{\Delta R^{\prime}}{R^{2}}=\dfrac{\Delta R_{1}}{R_{1}^{2}}+\dfrac{\Delta R_{2}}{R_{2}^{2}}$
$\Delta R^{\prime}=\left(R^{2}\right) \dfrac{\Delta R_{1}}{R_{1}^{2}}+\left(R^{2}\right) \dfrac{\Delta R_{2}}{R_{2}^{2}}$
$\left(\dfrac{66.7}{100}\right)^{2} 3+\left(\dfrac{66.7}{200}\right)^{2} 4$
$=1.8$
$R^{\prime}=66.7+1.8$ ohm.
Note:
In an electronic circuit, there are two fundamental ways to combine resistors in series (strung end to end) and in parallel (side by side). If a single resistor of that quantity is not available, a combination of resistors in an electric circuit can be used to provide the correct quantity of resistance in the circuit. So to get the required amount, we can use a different combination (series / parallel).
Formula used:
For series - $R={{R}_{1}}+{{R}_{2}}$
For Parallel - $R^{\prime}=\dfrac{R_{1} R_{1}}{R_{1}+R_{2}}$
Complete step by step solution:
(i) In a series resistor network, the individual resistors combine to give the series combination an equivalent resistance (RT). Without affecting the total resistance, current, or power of each resistor or circuit, the resistors in a series circuit can be interchanged. Resistance is directly proportional to the potential difference across a circuit's terminals. The potential difference is greater in sequences, thus offering more resistance.
The equivalent resistance of series
combination $R={{R}_{1}}+{{R}_{2}}=(100\pm 3)\text{ohm}+(200\pm 4)$ ohm
$=300 \pm 7$ ohm.
(ii) When two or more resistors are connected to each terminal of the other resistor or resistor so that both of their terminals are connected respectively, they are said to be connected together in parallel.
The equivalent resistance of parallel Combination
$R^{\prime}=\dfrac{R_{1} R_{1}}{R_{1}+R_{2}}=\dfrac{200}{3}=66.7$ ohm
Then, from $\dfrac{1}{R^{\prime}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}$
we get. $\dfrac{\Delta R^{\prime}}{R^{2}}=\dfrac{\Delta R_{1}}{R_{1}^{2}}+\dfrac{\Delta R_{2}}{R_{2}^{2}}$
$\Delta R^{\prime}=\left(R^{2}\right) \dfrac{\Delta R_{1}}{R_{1}^{2}}+\left(R^{2}\right) \dfrac{\Delta R_{2}}{R_{2}^{2}}$
$\left(\dfrac{66.7}{100}\right)^{2} 3+\left(\dfrac{66.7}{200}\right)^{2} 4$
$=1.8$
$R^{\prime}=66.7+1.8$ ohm.
Note:
In an electronic circuit, there are two fundamental ways to combine resistors in series (strung end to end) and in parallel (side by side). If a single resistor of that quantity is not available, a combination of resistors in an electric circuit can be used to provide the correct quantity of resistance in the circuit. So to get the required amount, we can use a different combination (series / parallel).
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