What steps did they do to get VEB to be 3 volts? Are you supposed to subtract VEB before taking the ln of both sides? I put my process so you can see where I got stuck. The image contains a mathematical derivation focused on determining the value of \( V_{EB} \), which is approximately 0.3 V. The steps are shown below:1. The initial equation is given as: \[ \frac{0.3 \, V}{10^{-14}} = V_{EB} + 10^{-4} e^{\frac{V_{EB}}{25 \, \text{mV}}} \]2. Simplifying the equation, the term \( \frac{10^{-14}}{10^{-14}} \) cancels out, resulting in: \[ 3 \times 10^{13} = V_{EB} + e^{\frac{V_{EB}}{26 \times 10^{-3}}} \]3. Taking the natural logarithm of both sides: \[ \ln(3 \times 10^{13}) = \ln\left(V_{EB} + e^{\frac{V_{EB}}{26 \times 10^{-3}}}\right) \]4. Highlighted analysis questions around the expression: \[ \ln\left(e^{\frac{V_{EB}}{26 \times 10^{-3}}}\right) \, ? \]5. The boxed conclusion from the derivation: \[ V_{EB} \approx 0.3 \, V \]The derivation process involves logarithmic transformations and simplifications of exponential equations, ultimately leading to determining the voltage \( V_{EB} \) with an approximation.

Answered: Image /qna-images/answer/0a402508-922c-414c-917c-c81761537dd7.jpg

Answered: VEB 25 mr V 0.3= VEB + VÓ VEB 0.3 V…

Similar questions

The figure below shows a simplified model of a cardiac defibrillator, a device used to resuscitate patients in ventricular fibrillation. S R Rorso + The capacitor C charge through the resistor R (when the switch S is toggled to left) and discharge current through the patient's torso which is of resistance (when the switch S is toggled to right). This phenomenon allows the heart's normal rhythm to be reestablished. (a) If the capacitor is initially uncharged with C = 7.0 µF; ɛ = 1240 V, find the value of R (in ohms) required to charge the capacitor to a voltage of 800 V in 2.1 s. (b) If the capacitor is then discharged across the patient's torso with Rtorso = 1200 Q, calculate the voltage (in V) across the capacitor after 4.0 ms.
Hall potentials are much larger for poor conductors than for good conductors. Why?
A resistor composed of carbon =(p=3.5*10^5 Ω*m) having a radius of 1.2mm and a length of 3.1mm is connected to a 6V batter. How much energy is dissipated by the resistor in 5 minutes?
A square loop has a charge density λ = |x| at y = ±a and [x]
F=0.8975r^-2.00 And re-writing it as: (4)F=0.8975/r^-2.00 We can now compare it with Coulomb’s equation: (1)F=K|q1q2|/r^2 So, comparing these equations (4) and (1) we realize that: k=|q1q2|=0.8975 Plugging in values for charges: q1 = -10µC, and q2 = 10µC, isolating k (Coulomb’s constant), and recalling that 1µC = 10-6 C, we get: k|-10µC*10µC|=0.8975 k=0.8934/|-10µC*10µC|= 0.8934/10*10^-6C*10*10^-6C (3) Use this Example calculation above, and the trendline equation (3) coefficients that you have obtained on your graph, to find k. Show your work below. Be careful with powers of ten. k= 0.8975/|-10µC*10µC|= 0.8975/10*10^-6C*10*10^-6C = (5) Now, important part in reporting the answer is properly showing its units. To derive the SI units for Coulomb’s constant k, let’s look again at the equation (1), and re-write it as: k=F*r^2/|q1q2| (6) Recalling the units for all parameters in this equation: Force, F – Newtons, [N] Distance, r –…
n 1 to 2 sentences, expiain why electromagnetc radiaHon can be modeled as b0th a wave and a partice.
A proton travels clockwise in the plane of the page under the influence of a magnetic field of 3T a) If the particle has an energy of 10-15J, find the radius of the circle b) What is the direction of the field? c) A long wire carries a current of 10A. A closed loop, in the form of a rectangle with the dimensions in the figure carries a current of 4 A. The side of the loop nearest the wire is 1mm from the wire. Find the force on the loop.

VEB 25 mr V 0.3= VEB + VÓ VEB 0.3 V l014 UEB/ 26x10-3 e 3x 1013= UEB t VEB/26 x/0-3 In(3x10"): In (UEB +(e e) VEB/26x03