# 物理代写|电动力学代写electromagnetism代考|PHYC20014

## 物理代写|电动力学代写electromagnetism代考|KIRCHHOFf’s Voltage and Current LaWS

While we are considering electroconductive fields, let us take some time to examine Kirchhoff’s laws which are more often treated as ‘circuit’ laws. Figure 4.6a shows a circuit in which two resistors are connected in series. This combination is connected to a source of potential V. Now, what happens as we move a positive test charge from the negative terminal to the positive terminal via the external circuit?

Let us move the test charge from the negative terminal through $R_2$ to junction between the two resistors. As there is an $E$ field set up in resistor $R_2$ we have to do work against the field. The magnitude of the $E$ field can be written as
$$E_2=\frac{V_2}{l_2}$$
where $l_2$ is the length of resistor $R_2$. Now, the force on our unit test charge is
\begin{aligned} F_2=& \frac{V_2}{l_2} \times 1 \ =& \frac{V_2}{l_2} \mathrm{~N} \end{aligned}
and so the work done in moving through the resistor is \begin{aligned} \text { Work done } &=\frac{V_2}{l_2} \times l_2 \ &=V_2 \end{aligned}
Similarly, the work done in moving through resistor $R_1$ is
\begin{aligned} \text { Work done } &=\frac{V_2}{l_2} \times l_1 \ &=V_1 \end{aligned}
Hence, the total work done in moving the test charge around the external circuit is
Total work done $=V_1+V_2$

## 物理代写|电动力学代写electromagnetism代考|Combinations of Resistors

Like capacitors, resistors come in standard values. Sometimes this can be very inconvenient, and so we have to make our own values. One way to do this is to trim an existing resistor by gently filing away at the body! This requires a very steady hand and a great deal of patience. An alternative is to combine resistors in series or parallel.
Figure 4.7a shows two resistors, $R_1$, and $R_2$, in parallel. We require to find the equivalent resistance of this arrangement. Let us connect a d.c. source, $V_s$, to the resistors. Now, both resistors have the same voltage across them, but they will pass different currents. So,
$$I_1=\frac{V_s}{R_1}$$
and
$$I_2=\frac{V_s}{R_2}$$
If we replace the two resistors by a single equivalent one of value $R_l$, the current taken by the new resistor, $I_r$, must be the same as that taken by parallel combination. Thus,
$$I_t=\frac{V_s}{R_t}$$
To be equivalent, $I_t$ must be the sum of the individual currents. So,
$$I_t=I_1+I_2$$
or
$$\frac{V_s}{R_t}=\frac{V_s}{R_1}+\frac{V_s}{R_2}$$

## 物理代写|电动力学代写electromagnetism代考|KIRCHHOFf’s Voltage and Current LaWS

$$E_2=\frac{V_2}{l_2}$$

$$F_2=\frac{V_2}{l_2} \times 1=\frac{V_2}{l_2} \mathrm{~N}$$

$$\text { Work done }=\frac{V_2}{l_2} \times l_2 \quad=V_2$$

$$\text { Work done }=\frac{V_2}{l_2} \times l_1 \quad=V_1$$

## 物理代写|电动力学代写electromagnetism代考|Combinations of Resistors

$$I_1=\frac{V_s}{R_1}$$

$$I_2=\frac{V_s}{R_2}$$

$$I_t=\frac{V_s}{R_t}$$

$$I_t=I_1+I_2$$

$$\frac{V_s}{R_t}=\frac{V_s}{R_1}+\frac{V_s}{R_2}$$

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