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Not quite - you want the cold sink to be as cold as possible. The cold sink is the cold air (or water or coolant) flowing over the heatsink. The heatsink is designed to efficiently connect the chip to the cold sink, so it needs to be as low resistance as possible. In thermodynamic calculations consider the chip==heatsink and the airflow to be the cold ...

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Think of it in terms of current, V, W, and Z are in series, so each is equally bright. X and Y are in parallel, so each gets half the current of the others. If you assume each bulb is a constant resistance R (not true for incandescent bulbs, by the way), then V,W and Z will each dissipate $i^2R$. For X and Y, since each has a current i/2, the power will be ...

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Current is produced in a metal when the free electrons in the metals acquire a drift velocity due to an electric field. But when these free electrons travel through the metal, their path is hindered by other atoms and particles and their electomagnetic pull. More the resistance, higher is this hindrance and lesser is the drift velocity. Hence, the current ...

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The answer depends on the circumstances: how do you change the resistance? Both the drift velocity and the number of available charge carriers can be changed. In a basic Drude model for electrical transport both, $n$, the charge carrier density and $\tau$, the time between collisions determine the resistance: $$\mathbf{J} = \left( \frac{n q^2 \tau}{m} ... 1 Your understanding is correct. From V=IR if voltage stays the same while resistance is increased, the current should be decreased. But if you have heard of another equation I = \frac{Q}{t} If current(I) is increased and the charge Q is fixed (Charge is fixed if the power supply is from power cells like battery). Time will be decreased. Which means ... 0 Let the battery voltage be V_S. Then, the battery current is$$I_S = \frac{V_S}{(100 + 100||100)\Omega} = \frac{V_S}{150 \Omega} $$The current through the series (left-most) resistor is just I_S and the current through each parallel resistor is just I_S/2. Since the resistor power is given by$$p_R = R \cdot I^2_R$$the power delivered to the ... 0 If the three resistors are connected in series, the total effective resistance is 3\times 2\Omega=6\Omega and the current is I=U/R_{\rm tot}=12/6{\rm A}=2{\rm A}, and the potential across each resistor is IR=2\times 2{\rm V}=4{\rm V}. However, an easy way to do it is, since we know in this case the current through each resistor is the same, so the ... 0 Yes this is common practice for stepper motors only that the strategy is a little bit different. Instead of the voltage value one controls the time average of the voltage through a pulse-width modulation. This works because the coil works like an integrator:$$ i_L(t) = i_0+\frac1{L}\int_{0}^t v_L\left(\bar t\right) d\bar t $$v_L would be$$ v_L(t) = ...

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The voltage across $R_1$ and $R_2$ will be the same, $V$. You are right that the net resistance will decrease to $(1/R_1+1/R_2)^{-1}$ and this change will be compensated by an increase in the current $I$ (Ohm's Law). $V$ will stay constant. Why? Because each component in a parallel circuit has two common nodes with each of the other components in the ...

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One finds a full explanation of this formula and much more about the connection between random walks and electric networks in the little gem Random walks and electric networks by Peter G. Doyle and J. Laurie Snell, freely available.

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See here. There's instructions on how to calcuate $V_{th}$ and $R_{th}$. Let $I_3$ be the current through $R_3$, $I_1$ through $R_1$, etc. With $V_{AB}$ open, by KVL, we have $V_1 - R_1I_1 - R_3 I_3 = 0$ and $V_1 - R_1 I_1 - R_2 I_2 -V_{AB} = 0$, but when $V_{AB}$ is open we have $I_2 = 0$ and so $I_1 = I_3$, so $V_1 = (R_1 + R_3) ... 1 Both examples are using the same methods. In the first example, the break in the circuit removes the 6 ohm and 5 ohm resistors altogether. All of the current then flows through the 2 ohm resistor, causing the Thevenin resistance to be 2 ohms. In the second example, the current first passes through$R_2$, then is split between$R_1$and$R_3$in parallel. ... 0 See at bottom. A formula for unbalanced Wheatstone bridge. (Lazy as shikamaru to write the stuff in latex). sorry for quality The book moved while scanning. 0 No, potential difference across each resistor need not be different always. In series, you should remember that potential difference across any resistor is proportional to its resistance. In other words, potential difference can be same across each resistor if each resistors value is same. If resistance values are different, say$R_1$and$R_2$, then by ... 0 Your reasoning is correct. If the solar cell is modelled as a voltage source$E$in series with an internal resistance$r$and the cell is connected to a load resistance$R_L$, the series current is given by Ohm's law: $$I = \frac{E}{r + R_L}$$ or $$E = (r + R_L)\cdot I$$ The output voltage$V$of the solar cell is the voltage across the load ... 0 You need to assume how the current gets divided across the whole circuit. For example, take the current coming out of the main battery as$I$, then when it reaches the loop that you have selected, it breaks into$I_1$and$I-I_1$. Do the same for all the loops, then apply Kirchoff's Rule. For capacitors, apply it exactly the same way you do it for batteries. ... 2 There are basically two kinds of addition here (ordinary addition and inverse-sum-inverse), representing series and parallel arrangements. You can represent the thing as a tree with alternating nodes of addition and ISI layers. The thing resolves pretty much down to a tree with N leaves. The magic is dealt with here http://oeis.org/A000669 . It talks ... 0 Air resistance acts only on macroscopic objects travelling through air. Electric fields are not affected by this at all. Electromagnetic waves do not even require a medium to propagate, so there is no point in even discussing how they can propagate through air. I think you are confusing Permittivity of air with air resistance, as anna mentioned in the ... 0 but it isn't the case with resistors in parallels (1) parallel connected resistors have identical voltage across $$V_{R1}=V_{R2}$$ (2) the voltage across a resistor is given by Ohm's law$V_R = I_R\cdot R\$. Applying Ohm's law to parallel connected resistors yields $$I_{R1}\cdot R_1 = I_{R2}\cdot R_2$$ Thus, conclude that the parallel connected ...

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